{"id":1195,"date":"2021-06-30T17:02:08","date_gmt":"2021-06-30T17:02:08","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/calculus-of-hyperbolic-and-inverse-hyperbolic-functions\/"},"modified":"2022-03-19T03:49:03","modified_gmt":"2022-03-19T03:49:03","slug":"calculus-of-hyperbolic-and-inverse-hyperbolic-functions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/calculus-of-hyperbolic-and-inverse-hyperbolic-functions\/","title":{"raw":"Calculus of Hyperbolic and Inverse Hyperbolic Functions","rendered":"Calculus of Hyperbolic and Inverse Hyperbolic Functions"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Apply the formulas for derivatives and integrals of the hyperbolic functions<\/li>\r\n \t<li>Apply the formulas for the derivatives of the inverse hyperbolic functions and their associated integrals<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>Derivatives and Integrals of the Hyperbolic Functions<\/h2>\r\n<p id=\"fs-id1167793637684\">Recall that the hyperbolic sine and hyperbolic cosine are defined as<\/p>\r\n\r\n<div id=\"fs-id1167793288534\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\text{sinh}x=\\dfrac{{e}^{x}-{e}^{\\text{\u2212}x}}{2}\\text{ and }\\text{cosh}x=\\dfrac{{e}^{x}+{e}^{\\text{\u2212}x}}{2}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167793233791\">The other hyperbolic functions are then defined in terms of [latex]\\text{sinh}x[\/latex] and [latex]\\text{cosh}x.[\/latex] The graphs of the hyperbolic functions are shown in the following figure.<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"958\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11213342\/CNX_Calc_Figure_06_09_001.jpg\" alt=\"This figure has six graphs. The first graph labeled \u201ca\u201d is of the function y=sinh(x). It is an increasing function from the 3rd quadrant, through the origin to the first quadrant. The second graph is labeled \u201cb\u201d and is of the function y=cosh(x). It decreases in the second quadrant to the intercept y=1, then becomes an increasing function. The third graph labeled \u201cc\u201d is of the function y=tanh(x). It is an increasing function from the third quadrant, through the origin, to the first quadrant. The fourth graph is labeled \u201cd\u201d and is of the function y=coth(x). It has two pieces, one in the third quadrant and one in the first quadrant with a vertical asymptote at the y-axis. The fifth graph is labeled \u201ce\u201d and is of the function y=sech(x). It is a curve above the x-axis, increasing in the second quadrant, to the y-axis at y=1 and then decreases. The sixth graph is labeled \u201cf\u201d and is of the function y=csch(x). It has two pieces, one in the third quadrant and one in the first quadrant with a vertical asymptote at the y-axis.\" width=\"958\" height=\"749\" \/> Figure 1. Graphs of the hyperbolic functions.[\/caption]\r\n<p id=\"fs-id1167793925306\">It is easy to develop differentiation formulas for the hyperbolic functions. For example, looking at [latex]\\text{sinh}x[\/latex] we have<\/p>\r\n\r\n<div id=\"fs-id1167794049237\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill \\frac{d}{dx}(\\text{sinh}x)&amp; =\\frac{d}{dx}(\\frac{{e}^{x}-{e}^{\\text{\u2212}x}}{2})\\hfill \\\\ &amp; =\\frac{1}{2}\\left[\\frac{d}{dx}({e}^{x})-\\frac{d}{dx}({e}^{\\text{\u2212}x})\\right]\\hfill \\\\ &amp; =\\frac{1}{2}\\left[{e}^{x}+{e}^{\\text{\u2212}x}\\right]=\\text{cosh}x.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167793278441\">Similarly, [latex](\\frac{d}{dx})\\text{cosh}x=\\text{sinh}x.[\/latex] We summarize the differentiation formulas for the hyperbolic functions in the following table.<\/p>\r\n\r\n<table id=\"fs-id1167793363804\" summary=\"This is a table of two columns. The first column is labeled f(x). Its entries are the hyperbolic trigonometric functions. The second column is labeled d\/dx f(x) and is the corresponding derivatives of the hyperbolic trigonometric functions.\"><caption>Derivatives of the Hyperbolic Functions<\/caption>\r\n<thead>\r\n<tr valign=\"top\">\r\n<th>[latex]f(x)[\/latex]<\/th>\r\n<th>[latex]\\frac{d}{dx}f(x)[\/latex]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td>[latex]\\text{sinh}x[\/latex]<\/td>\r\n<td>[latex]\\text{cosh}x[\/latex]<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex]\\text{cosh}x[\/latex]<\/td>\r\n<td>[latex]\\text{sinh}x[\/latex]<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex]\\text{tanh}x[\/latex]<\/td>\r\n<td>[latex]{\\text{sech}}^{2}x[\/latex]<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex]\\text{coth}x[\/latex]<\/td>\r\n<td>[latex]\\text{\u2212}{\\text{csch}}^{2}x[\/latex]<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex]\\text{sech}x[\/latex]<\/td>\r\n<td>[latex]\\text{\u2212}\\text{sech}x\\text{tanh}x[\/latex]<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex]\\text{csch}x[\/latex]<\/td>\r\n<td>[latex]\\text{\u2212}\\text{csch}x\\text{coth}x[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"fs-id1167794139383\">Let\u2019s take a moment to compare the derivatives of the hyperbolic functions with the derivatives of the standard trigonometric functions. There are a lot of similarities, but differences as well. For example, the derivatives of the sine functions match: [latex](\\frac{d}{dx}) \\sin x= \\cos x[\/latex] and [latex](\\frac{d}{dx})\\text{sinh}x=\\text{cosh}x.[\/latex] The derivatives of the cosine functions, however, differ in sign: [latex](\\frac{d}{dx}) \\cos x=\\text{\u2212} \\sin x,[\/latex] but [latex](\\frac{d}{dx})\\text{cosh}x=\\text{sinh}x.[\/latex] As we continue our examination of the hyperbolic functions, we must be mindful of their similarities and differences to the standard trigonometric functions.<\/p>\r\n<p id=\"fs-id1167794334421\">These differentiation formulas for the hyperbolic functions lead directly to the following integral formulas.<\/p>\r\n\r\n<div id=\"fs-id1167793944458\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cccccccc}\\hfill \\displaystyle\\int \\text{sinh}udu&amp; =\\hfill &amp; \\text{cosh}u+C\\hfill &amp; &amp; &amp; \\hfill \\displaystyle\\int {\\text{csch}}^{2}udu&amp; =\\hfill &amp; \\text{\u2212}\\text{coth}u+C\\hfill \\\\ \\hfill \\displaystyle\\int \\text{cosh}udu&amp; =\\hfill &amp; \\text{sinh}u+C\\hfill &amp; &amp; &amp; \\hfill \\displaystyle\\int \\text{sech}u\\text{tanh}udu&amp; =\\hfill &amp; \\text{\u2212}\\text{sech}u+C\\hfill \\\\ \\hfill \\displaystyle\\int {\\text{sech}}^{2}udu&amp; =\\hfill &amp; \\text{tanh}u+C\\hfill &amp; &amp; &amp; \\hfill \\displaystyle\\int \\text{csch}u\\text{coth}udu&amp; =\\hfill &amp; \\text{\u2212}\\text{csch}u+C\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<div id=\"fs-id1167793852238\" class=\"textbook exercises\">\r\n<h3>Example: Differentiating Hyperbolic Functions<\/h3>\r\n<p id=\"fs-id1167793278457\">Evaluate the following derivatives:<\/p>\r\n\r\n<ol id=\"fs-id1167794060382\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]\\frac{d}{dx}(\\text{sinh}({x}^{2}))[\/latex]<\/li>\r\n \t<li>[latex]\\frac{d}{dx}{(\\text{cosh}x)}^{2}[\/latex]<\/li>\r\n<\/ol>\r\n<div id=\"fs-id1167793936071\" class=\"exercise\">[reveal-answer q=\"fs-id1167794098810\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167794098810\"]\r\n<p id=\"fs-id1167794098810\">Using the formulas in the table on derivatives\u00a0 of the hyperbolic functions and the chain rule, we get<\/p>\r\n\r\n<ol id=\"fs-id1167793500488\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]\\frac{d}{dx}(\\text{sinh}({x}^{2}))=\\text{cosh}({x}^{2})\u00b72x[\/latex]<\/li>\r\n \t<li>[latex]\\frac{d}{dx}{(\\text{cosh}x)}^{2}=2\\text{cosh}x\\text{sinh}x[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167793970793\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1167794028505\">Evaluate the following derivatives:<\/p>\r\n\r\n<ol id=\"fs-id1167794043812\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]\\frac{d}{dx}(\\text{tanh}({x}^{2}+3x))[\/latex]<\/li>\r\n \t<li>[latex]\\frac{d}{dx}(\\dfrac{1}{{(\\text{sinh}x)}^{2}})[\/latex]<\/li>\r\n<\/ol>\r\n<div>[reveal-answer q=\"585894\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"585894\"]Use the formulas in the last example and apply the chain rule as necessary.[\/hidden-answer]<\/div>\r\n<div><\/div>\r\n<div id=\"fs-id1167794145876\" class=\"exercise\">[reveal-answer q=\"fs-id1167794187159\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167794187159\"]\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]\\frac{d}{dx}(\\text{tanh}({x}^{2}+3x))=({\\text{sech}}^{2}({x}^{2}+3x))(2x+3)[\/latex]<\/li>\r\n \t<li>[latex]\\frac{d}{dx}(\\frac{1}{{(\\text{sinh}x)}^{2}})=\\frac{d}{dx}{(\\text{sinh}x)}^{-2}=-2{(\\text{sinh}x)}^{-3}\\text{cosh}x[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\nWatch the following video to see the worked solution to Example: Differentiating Hyperbolic Functions and the above Try It.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/ozRCEFk1tmA?controls=0&amp;start=134&amp;end=300&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>[reveal-answer q=\"266834\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266834\"]For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/2.9CalculusOfHyperbolicFunctions134to300_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"2.9 Calculus of Hyperbolic Functions\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<div id=\"fs-id1167793279424\" class=\"textbook exercises\">\r\n<h3>example: Integrals Involving Hyperbolic Functions<\/h3>\r\n<p id=\"fs-id1167793965336\">Evaluate the following integrals:<\/p>\r\n\r\n<ol id=\"fs-id1167794050806\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]\\displaystyle\\int x\\text{cosh}({x}^{2})dx[\/latex]<\/li>\r\n \t<li>[latex]\\displaystyle\\int \\text{tanh}xdx[\/latex]<\/li>\r\n<\/ol>\r\n<div id=\"fs-id1167794051555\" class=\"exercise\">[reveal-answer q=\"fs-id1167793254882\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167793254882\"]\r\n<p id=\"fs-id1167793254882\">We can use [latex]u[\/latex]-substitution in both cases.<\/p>\r\n\r\n<ol id=\"fs-id1167794042760\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>Let [latex]u={x}^{2}.[\/latex] Then, [latex]du=2xdx[\/latex] and\r\n<div id=\"fs-id1167794337482\" class=\"equation unnumbered\">[latex]\\displaystyle\\int x\\text{cosh}({x}^{2})dx=\\displaystyle\\int \\frac{1}{2}\\text{cosh}udu=\\frac{1}{2}\\text{sinh}u+C=\\frac{1}{2}\\text{sinh}({x}^{2})+C.[\/latex]<\/div><\/li>\r\n \t<li>Let [latex]u=\\text{cosh}x.[\/latex] Then, [latex]du=\\text{sinh}xdx[\/latex] and\r\n<div id=\"fs-id1167794043265\" class=\"equation unnumbered\">[latex]\\displaystyle\\int \\text{tanh}xdx=\\displaystyle\\int \\frac{\\text{sinh}x}{\\text{cosh}x}dx=\\displaystyle\\int \\frac{1}{u}du=\\text{ln}|u|+C=\\text{ln}|\\text{cosh}x|+C.[\/latex]<\/div>\r\nNote that [latex]\\text{cosh}x&gt;0[\/latex] for all [latex]x,[\/latex] so we can eliminate the absolute value signs and obtain\r\n<div id=\"fs-id1167793358384\" class=\"equation unnumbered\">[latex]\\displaystyle\\int \\text{tanh}xdx=\\text{ln}(\\text{cosh}x)+C.[\/latex]<\/div><\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167793504384\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1167793819956\">Evaluate the following integrals:<\/p>\r\n\r\n<ol id=\"fs-id1167793985861\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]\\displaystyle\\int {\\text{sinh}}^{3}x\\text{cosh}xdx[\/latex]<\/li>\r\n \t<li>[latex]\\displaystyle\\int {\\text{sech}}^{2}(3x)dx[\/latex]<\/li>\r\n<\/ol>\r\n<div>[reveal-answer q=\"385579\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"385579\"]Use the formulas above and apply [latex]u[\/latex]-substitution as necessary.[\/hidden-answer]<\/div>\r\n<div><\/div>\r\n<div id=\"fs-id1167794146347\" class=\"exercise\">[reveal-answer q=\"fs-id1167793373829\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167793373829\"]\r\n<ol id=\"fs-id1167793373829\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]\\displaystyle\\int {\\text{sinh}}^{3}x\\text{cosh}xdx=\\frac{{\\text{sinh}}^{4}x}{4}+C[\/latex]<\/li>\r\n \t<li>[latex]\\displaystyle\\int {\\text{sech}}^{2}(3x)dx=\\frac{\\text{tanh}(3x)}{3}+C[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\nWatch the following video to see the worked solution to the above Try It.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/ozRCEFk1tmA?controls=0&amp;start=478&amp;end=598&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>[reveal-answer q=\"266833\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266833\"]For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/2.9CalculusOfHyperbolicFunctions478to598_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"2.9 Calculus of Hyperbolic Functions\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]223494[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Calculus of Inverse Hyperbolic Functions<\/h2>\r\nLooking at the graphs of the hyperbolic functions, we see that with appropriate range restrictions, they all have inverses. Most of the necessary range restrictions can be discerned by close examination of the graphs. The domains and ranges of the inverse hyperbolic functions are summarized in the following table.\r\n<table id=\"fs-id1167793362111\" summary=\"This table has three columns. The first column is labeled function and has the inverse hyperbolic functions listed in the column. The second column is labeled domain and has the domains of the inverse hyperbolic functions. The third column is labeled range and has the ranges of the inverse hyperbolic functions.\"><caption>Domains and Ranges of the Inverse Hyperbolic Functions<\/caption>\r\n<thead>\r\n<tr valign=\"top\">\r\n<th>Function<\/th>\r\n<th>Domain<\/th>\r\n<th>Range<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td>[latex]{\\text{sinh}}^{-1}x[\/latex]<\/td>\r\n<td>[latex](\\text{\u2212}\\infty ,\\infty )[\/latex]<\/td>\r\n<td>[latex](\\text{\u2212}\\infty ,\\infty )[\/latex]<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex]{\\text{cosh}}^{-1}x[\/latex]<\/td>\r\n<td>[latex](1,\\infty )[\/latex]<\/td>\r\n<td>[latex][0,\\infty )[\/latex]<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex]{\\text{tanh}}^{-1}x[\/latex]<\/td>\r\n<td>[latex](-1,1)[\/latex]<\/td>\r\n<td>[latex](\\text{\u2212}\\infty ,\\infty )[\/latex]<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex]{\\text{coth}}^{-1}x[\/latex]<\/td>\r\n<td>[latex](\\text{\u2212}\\infty ,-1)\\cup (1,\\infty )[\/latex]<\/td>\r\n<td>[latex](\\text{\u2212}\\infty ,0)\\cup (0,\\infty )[\/latex]<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex]{\\text{sech}}^{-1}x[\/latex]<\/td>\r\n<td>[latex](0\\text{, 1})[\/latex]<\/td>\r\n<td>[latex][0,\\infty )[\/latex]<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex]{\\text{csch}}^{-1}x[\/latex]<\/td>\r\n<td>[latex](\\text{\u2212}\\infty ,0)\\cup (0,\\infty )[\/latex]<\/td>\r\n<td>[latex](\\text{\u2212}\\infty ,0)\\cup (0,\\infty )[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"fs-id1167793998256\">The graphs of the inverse hyperbolic functions are shown in the following figure.<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"958\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11213346\/CNX_Calc_Figure_06_09_002.jpg\" alt=\"This figure has six graphs. The first graph labeled \u201ca\u201d is of the function y=sinh^-1(x). It is an increasing function from the 3rd quadrant, through the origin to the first quadrant. The second graph is labeled \u201cb\u201d and is of the function y=cosh^-1(x). It is in the first quadrant, beginning on the x-axis at 2 and increasing. The third graph labeled \u201cc\u201d is of the function y=tanh^-1(x). It is an increasing function from the third quadrant, through the origin, to the first quadrant. The fourth graph is labeled \u201cd\u201d and is of the function y=coth^-1(x). It has two pieces, one in the third quadrant and one in the first quadrant with a vertical asymptote at the y-axis. The fifth graph is labeled \u201ce\u201d and is of the function y=sech^-1(x). It is a curve decreasing in the first quadrant and stopping on the x-axis at x=1. The sixth graph is labeled \u201cf\u201d and is of the function y=csch^-1(x). It has two pieces, one in the third quadrant and one in the first quadrant with a vertical asymptote at the y-axis.\" width=\"958\" height=\"749\" \/> Figure 2. Graphs of the inverse hyperbolic functions.[\/caption]\r\n<p id=\"fs-id1167793259261\">To find the derivatives of the inverse functions, we use implicit differentiation. We have<\/p>\r\n\r\n<div id=\"fs-id1167793959214\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill y&amp; =\\hfill &amp; {\\text{sinh}}^{-1}x\\hfill \\\\ \\hfill \\text{sinh}y&amp; =\\hfill &amp; x\\hfill \\\\ \\hfill \\frac{d}{dx}\\text{sinh}y&amp; =\\hfill &amp; \\frac{d}{dx}x\\hfill \\\\ \\hfill \\text{cosh}y\\frac{dy}{dx}&amp; =\\hfill &amp; 1.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167793589391\">Recall that [latex]{\\text{cosh}}^{2}y-{\\text{sinh}}^{2}y=1,[\/latex] so [latex]\\text{cosh}y=\\sqrt{1+{\\text{sinh}}^{2}y}.[\/latex] Then,<\/p>\r\n\r\n<div id=\"fs-id1167794042534\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{dy}{dx}=\\frac{1}{\\text{cosh}y}=\\frac{1}{\\sqrt{1+{\\text{sinh}}^{2}y}}=\\frac{1}{\\sqrt{1+{x}^{2}}}.[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167793638871\">We can derive differentiation formulas for the other inverse hyperbolic functions in a similar fashion. These differentiation formulas are summarized in the following table.<\/p>\r\n\r\n<table id=\"fs-id1167794020892\" summary=\"This table has two columns. The first column is labeled f(x) and has the inverse hyperbolic functions as entries. The second column is labeled d\/dx f(x) and is the derivatives of the inverse hyperbolic functions.\"><caption>Derivatives of the Inverse Hyperbolic Functions<\/caption>\r\n<thead>\r\n<tr valign=\"top\">\r\n<th>[latex]f(x)[\/latex]<\/th>\r\n<th>[latex]\\frac{d}{dx}f(x)[\/latex]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td>[latex]{\\text{sinh}}^{-1}x[\/latex]<\/td>\r\n<td>[latex]\\frac{1}{\\sqrt{1+{x}^{2}}}[\/latex]<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex]{\\text{cosh}}^{-1}x[\/latex]<\/td>\r\n<td>[latex]\\frac{1}{\\sqrt{{x}^{2}-1}}[\/latex]<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex]{\\text{tanh}}^{-1}x[\/latex]<\/td>\r\n<td>[latex]\\frac{1}{1-{x}^{2}}[\/latex]<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex]{\\text{coth}}^{-1}x[\/latex]<\/td>\r\n<td>[latex]\\frac{1}{1-{x}^{2}}[\/latex]<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex]{\\text{sech}}^{-1}x[\/latex]<\/td>\r\n<td>[latex]\\frac{-1}{x\\sqrt{1-{x}^{2}}}[\/latex]<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex]{\\text{csch}}^{-1}x[\/latex]<\/td>\r\n<td>[latex]\\frac{-1}{|x|\\sqrt{1+{x}^{2}}}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"fs-id1167794119173\">Note that the derivatives of [latex]{\\text{tanh}}^{-1}x[\/latex] and [latex]{\\text{coth}}^{-1}x[\/latex] are the same. Thus, when we integrate [latex]1\\text{\/}(1-{x}^{2}),[\/latex] we need to select the proper antiderivative based on the domain of the functions and the values of [latex]x.[\/latex] Integration formulas involving the inverse hyperbolic functions are summarized as follows.<\/p>\r\n\r\n<div id=\"fs-id1167793951927\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cccccccc}\\hfill \\displaystyle\\int \\frac{1}{\\sqrt{1+{u}^{2}}}du&amp; =\\hfill &amp; {\\text{sinh}}^{-1}u+C\\hfill &amp; &amp; &amp; \\hfill \\displaystyle\\int \\frac{1}{u\\sqrt{1-{u}^{2}}}du&amp; =\\hfill &amp; \\text{\u2212}{\\text{sech}}^{-1}|u|+C\\hfill \\\\ \\hfill \\displaystyle\\int \\frac{1}{\\sqrt{{u}^{2}-1}}du&amp; =\\hfill &amp; {\\text{cosh}}^{-1}u+C\\hfill &amp; &amp; &amp; \\hfill \\displaystyle\\int \\frac{1}{u\\sqrt{1+{u}^{2}}}du&amp; =\\hfill &amp; \\text{\u2212}{\\text{csch}}^{-1}|u|+C\\hfill \\\\ \\hfill \\displaystyle\\int \\frac{1}{1-{u}^{2}}du&amp; =\\hfill &amp; \\bigg\\{\\begin{array}{c}{\\text{tanh}}^{-1}u+C\\text{ if }|u|&lt;1\\hfill \\\\ {\\text{coth}}^{-1}u+C\\text{ if }|u|&gt;1\\hfill \\end{array}\\hfill &amp; &amp; &amp; &amp; &amp; \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<div id=\"fs-id1167793984979\" class=\"textbook exercises\">\r\n<h3>Example: Differentiating Inverse Hyperbolic Functions<\/h3>\r\n<p id=\"fs-id1167793372563\">Evaluate the following derivatives:<\/p>\r\n\r\n<ol id=\"fs-id1167794155431\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]\\frac{d}{dx}({\\text{sinh}}^{-1}(\\frac{x}{3}))[\/latex]<\/li>\r\n \t<li>[latex]\\frac{d}{dx}{({\\text{tanh}}^{-1}x)}^{2}[\/latex]<\/li>\r\n<\/ol>\r\n<div id=\"fs-id1167793984981\" class=\"exercise\">[reveal-answer q=\"fs-id1167793949787\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167793949787\"]\r\n<p id=\"fs-id1167793949787\">Using the formulas in the table on derivatives of the inverse hyperbolic functions and the chain rule, we obtain the following results:<\/p>\r\n\r\n<ol id=\"fs-id1167793298499\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]\\frac{d}{dx}({\\text{sinh}}^{-1}(\\frac{x}{3}))=\\frac{1}{3\\sqrt{1+\\frac{{x}^{2}}{9}}}=\\frac{1}{\\sqrt{9+{x}^{2}}}[\/latex]<\/li>\r\n \t<li>[latex]\\frac{d}{dx}{({\\text{tanh}}^{-1}x)}^{2}=\\frac{2({\\text{tanh}}^{-1}x)}{1-{x}^{2}}[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167793626577\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1167794332395\">Evaluate the following derivatives:<\/p>\r\n\r\n<ol id=\"fs-id1167793590727\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]\\frac{d}{dx}({\\text{cosh}}^{-1}(3x))[\/latex]<\/li>\r\n \t<li>[latex]\\frac{d}{dx}{({\\text{coth}}^{-1}x)}^{3}[\/latex]<\/li>\r\n<\/ol>\r\n<div>[reveal-answer q=\"358726\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"358726\"]Use the formulas in the table on derivatives of the inverse hyperbolic functions above and apply the chain rule as necessary.[\/hidden-answer]<\/div>\r\n<div><\/div>\r\n<div id=\"fs-id1167793626580\" class=\"exercise\">[reveal-answer q=\"fs-id1167793498614\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167793498614\"]\r\n<ol id=\"fs-id1167793498614\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]\\frac{d}{dx}({\\text{cosh}}^{-1}(3x))=\\frac{3}{\\sqrt{9{x}^{2}-1}}[\/latex]<\/li>\r\n \t<li>[latex]\\frac{d}{dx}{({\\text{coth}}^{-1}x)}^{3}=\\frac{3{({\\text{coth}}^{-1}x)}^{2}}{1-{x}^{2}}[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\nWatch the following video to see the worked solution to the above Try It.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/ozRCEFk1tmA?controls=0&amp;start=827&amp;end=907&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>[reveal-answer q=\"266836\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266836\"]For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/2.9CalculusOfHyperbolicFunctions827to907_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"2.9 Calculus of Hyperbolic Functions\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<div id=\"fs-id1167793285224\" class=\"textbook exercises\">\r\n<h3>Example: Integrals Involving Inverse Hyperbolic Functions<\/h3>\r\n<p id=\"fs-id1167793871641\">Evaluate the following integrals:<\/p>\r\n\r\n<ol id=\"fs-id1167793937415\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]\\displaystyle\\int \\frac{1}{\\sqrt{4{x}^{2}-1}}dx[\/latex]<\/li>\r\n \t<li>[latex]\\displaystyle\\int \\frac{1}{2x\\sqrt{1-9{x}^{2}}}dx[\/latex]<\/li>\r\n<\/ol>\r\n<div id=\"fs-id1167793285226\" class=\"exercise\">[reveal-answer q=\"fs-id1167793361818\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167793361818\"]\r\n<p id=\"fs-id1167793361818\">We can use [latex]u\\text{-substitution}[\/latex] in both cases.<\/p>\r\n\r\n<ol id=\"fs-id1167794296597\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>Let [latex]u=2x.[\/latex] Then, [latex]du=2dx[\/latex] and we have\r\n<div id=\"fs-id1167793883719\" class=\"equation unnumbered\">[latex]\\displaystyle\\int \\frac{1}{\\sqrt{4{x}^{2}-1}}dx=\\displaystyle\\int \\frac{1}{2\\sqrt{{u}^{2}-1}}du=\\frac{1}{2}{\\text{cosh}}^{-1}u+C=\\frac{1}{2}{\\text{cosh}}^{-1}(2x)+C.[\/latex]<\/div><\/li>\r\n \t<li>Let [latex]u=3x.[\/latex] Then, [latex]du=3dx[\/latex] and we obtain\r\n<div id=\"fs-id1167793444519\" class=\"equation unnumbered\">[latex]\\displaystyle\\int \\frac{1}{2x\\sqrt{1-9{x}^{2}}}dx=\\frac{1}{2}\\displaystyle\\int \\frac{1}{u\\sqrt{1-{u}^{2}}}du=-\\frac{1}{2}{\\text{sech}}^{-1}|u|+C=-\\frac{1}{2}{\\text{sech}}^{-1}|3x|+C.[\/latex]<\/div><\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167793925131\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1167793265985\">Evaluate the following integrals:<\/p>\r\n\r\n<ol id=\"fs-id1167793265989\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]\\displaystyle\\int \\frac{1}{\\sqrt{{x}^{2}-4}}dx,\\text{}x&gt;2[\/latex]<\/li>\r\n \t<li>[latex]\\displaystyle\\int \\frac{1}{\\sqrt{1-{e}^{2x}}}dx[\/latex]<\/li>\r\n<\/ol>\r\n<div>[reveal-answer q=\"351257\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"351257\"]Use the formulas above and apply [latex]u\\text{-substitution}[\/latex] as necessary.[\/hidden-answer]<\/div>\r\n<div><\/div>\r\n<div id=\"fs-id1167793969727\" class=\"exercise\">[reveal-answer q=\"fs-id1167793371557\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167793371557\"]\r\n<ol id=\"fs-id1167793371557\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]\\displaystyle\\int \\frac{1}{\\sqrt{{x}^{2}-4}}dx={\\text{cosh}}^{-1}(\\frac{x}{2})+C[\/latex]<\/li>\r\n \t<li>[latex]\\displaystyle\\int \\frac{1}{\\sqrt{1-{e}^{2x}}}dx=\\text{\u2212}{\\text{sech}}^{-1}({e}^{x})+C[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\nWatch the following video to see the worked solution to the above Try It.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/ozRCEFk1tmA?controls=0&amp;start=1140&amp;end=1263&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>[reveal-answer q=\"266837\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266837\"]For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/2.9CalculusOfHyperbolicFunctions1140to1263_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"2.9 Calculus of Hyperbolic Functions\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]223496[\/ohm_question]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Apply the formulas for derivatives and integrals of the hyperbolic functions<\/li>\n<li>Apply the formulas for the derivatives of the inverse hyperbolic functions and their associated integrals<\/li>\n<\/ul>\n<\/div>\n<h2>Derivatives and Integrals of the Hyperbolic Functions<\/h2>\n<p id=\"fs-id1167793637684\">Recall that the hyperbolic sine and hyperbolic cosine are defined as<\/p>\n<div id=\"fs-id1167793288534\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\text{sinh}x=\\dfrac{{e}^{x}-{e}^{\\text{\u2212}x}}{2}\\text{ and }\\text{cosh}x=\\dfrac{{e}^{x}+{e}^{\\text{\u2212}x}}{2}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167793233791\">The other hyperbolic functions are then defined in terms of [latex]\\text{sinh}x[\/latex] and [latex]\\text{cosh}x.[\/latex] The graphs of the hyperbolic functions are shown in the following figure.<\/p>\n<div style=\"width: 968px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11213342\/CNX_Calc_Figure_06_09_001.jpg\" alt=\"This figure has six graphs. The first graph labeled \u201ca\u201d is of the function y=sinh(x). It is an increasing function from the 3rd quadrant, through the origin to the first quadrant. The second graph is labeled \u201cb\u201d and is of the function y=cosh(x). It decreases in the second quadrant to the intercept y=1, then becomes an increasing function. The third graph labeled \u201cc\u201d is of the function y=tanh(x). It is an increasing function from the third quadrant, through the origin, to the first quadrant. The fourth graph is labeled \u201cd\u201d and is of the function y=coth(x). It has two pieces, one in the third quadrant and one in the first quadrant with a vertical asymptote at the y-axis. The fifth graph is labeled \u201ce\u201d and is of the function y=sech(x). It is a curve above the x-axis, increasing in the second quadrant, to the y-axis at y=1 and then decreases. The sixth graph is labeled \u201cf\u201d and is of the function y=csch(x). It has two pieces, one in the third quadrant and one in the first quadrant with a vertical asymptote at the y-axis.\" width=\"958\" height=\"749\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 1. Graphs of the hyperbolic functions.<\/p>\n<\/div>\n<p id=\"fs-id1167793925306\">It is easy to develop differentiation formulas for the hyperbolic functions. For example, looking at [latex]\\text{sinh}x[\/latex] we have<\/p>\n<div id=\"fs-id1167794049237\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill \\frac{d}{dx}(\\text{sinh}x)& =\\frac{d}{dx}(\\frac{{e}^{x}-{e}^{\\text{\u2212}x}}{2})\\hfill \\\\ & =\\frac{1}{2}\\left[\\frac{d}{dx}({e}^{x})-\\frac{d}{dx}({e}^{\\text{\u2212}x})\\right]\\hfill \\\\ & =\\frac{1}{2}\\left[{e}^{x}+{e}^{\\text{\u2212}x}\\right]=\\text{cosh}x.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167793278441\">Similarly, [latex](\\frac{d}{dx})\\text{cosh}x=\\text{sinh}x.[\/latex] We summarize the differentiation formulas for the hyperbolic functions in the following table.<\/p>\n<table id=\"fs-id1167793363804\" summary=\"This is a table of two columns. The first column is labeled f(x). Its entries are the hyperbolic trigonometric functions. The second column is labeled d\/dx f(x) and is the corresponding derivatives of the hyperbolic trigonometric functions.\">\n<caption>Derivatives of the Hyperbolic Functions<\/caption>\n<thead>\n<tr valign=\"top\">\n<th>[latex]f(x)[\/latex]<\/th>\n<th>[latex]\\frac{d}{dx}f(x)[\/latex]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td>[latex]\\text{sinh}x[\/latex]<\/td>\n<td>[latex]\\text{cosh}x[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]\\text{cosh}x[\/latex]<\/td>\n<td>[latex]\\text{sinh}x[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]\\text{tanh}x[\/latex]<\/td>\n<td>[latex]{\\text{sech}}^{2}x[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]\\text{coth}x[\/latex]<\/td>\n<td>[latex]\\text{\u2212}{\\text{csch}}^{2}x[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]\\text{sech}x[\/latex]<\/td>\n<td>[latex]\\text{\u2212}\\text{sech}x\\text{tanh}x[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]\\text{csch}x[\/latex]<\/td>\n<td>[latex]\\text{\u2212}\\text{csch}x\\text{coth}x[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-id1167794139383\">Let\u2019s take a moment to compare the derivatives of the hyperbolic functions with the derivatives of the standard trigonometric functions. There are a lot of similarities, but differences as well. For example, the derivatives of the sine functions match: [latex](\\frac{d}{dx}) \\sin x= \\cos x[\/latex] and [latex](\\frac{d}{dx})\\text{sinh}x=\\text{cosh}x.[\/latex] The derivatives of the cosine functions, however, differ in sign: [latex](\\frac{d}{dx}) \\cos x=\\text{\u2212} \\sin x,[\/latex] but [latex](\\frac{d}{dx})\\text{cosh}x=\\text{sinh}x.[\/latex] As we continue our examination of the hyperbolic functions, we must be mindful of their similarities and differences to the standard trigonometric functions.<\/p>\n<p id=\"fs-id1167794334421\">These differentiation formulas for the hyperbolic functions lead directly to the following integral formulas.<\/p>\n<div id=\"fs-id1167793944458\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cccccccc}\\hfill \\displaystyle\\int \\text{sinh}udu& =\\hfill & \\text{cosh}u+C\\hfill & & & \\hfill \\displaystyle\\int {\\text{csch}}^{2}udu& =\\hfill & \\text{\u2212}\\text{coth}u+C\\hfill \\\\ \\hfill \\displaystyle\\int \\text{cosh}udu& =\\hfill & \\text{sinh}u+C\\hfill & & & \\hfill \\displaystyle\\int \\text{sech}u\\text{tanh}udu& =\\hfill & \\text{\u2212}\\text{sech}u+C\\hfill \\\\ \\hfill \\displaystyle\\int {\\text{sech}}^{2}udu& =\\hfill & \\text{tanh}u+C\\hfill & & & \\hfill \\displaystyle\\int \\text{csch}u\\text{coth}udu& =\\hfill & \\text{\u2212}\\text{csch}u+C\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<div id=\"fs-id1167793852238\" class=\"textbook exercises\">\n<h3>Example: Differentiating Hyperbolic Functions<\/h3>\n<p id=\"fs-id1167793278457\">Evaluate the following derivatives:<\/p>\n<ol id=\"fs-id1167794060382\" style=\"list-style-type: lower-alpha;\">\n<li>[latex]\\frac{d}{dx}(\\text{sinh}({x}^{2}))[\/latex]<\/li>\n<li>[latex]\\frac{d}{dx}{(\\text{cosh}x)}^{2}[\/latex]<\/li>\n<\/ol>\n<div id=\"fs-id1167793936071\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167794098810\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167794098810\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167794098810\">Using the formulas in the table on derivatives\u00a0 of the hyperbolic functions and the chain rule, we get<\/p>\n<ol id=\"fs-id1167793500488\" style=\"list-style-type: lower-alpha;\">\n<li>[latex]\\frac{d}{dx}(\\text{sinh}({x}^{2}))=\\text{cosh}({x}^{2})\u00b72x[\/latex]<\/li>\n<li>[latex]\\frac{d}{dx}{(\\text{cosh}x)}^{2}=2\\text{cosh}x\\text{sinh}x[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1167793970793\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1167794028505\">Evaluate the following derivatives:<\/p>\n<ol id=\"fs-id1167794043812\" style=\"list-style-type: lower-alpha;\">\n<li>[latex]\\frac{d}{dx}(\\text{tanh}({x}^{2}+3x))[\/latex]<\/li>\n<li>[latex]\\frac{d}{dx}(\\dfrac{1}{{(\\text{sinh}x)}^{2}})[\/latex]<\/li>\n<\/ol>\n<div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q585894\">Hint<\/span><\/p>\n<div id=\"q585894\" class=\"hidden-answer\" style=\"display: none\">Use the formulas in the last example and apply the chain rule as necessary.<\/div>\n<\/div>\n<\/div>\n<div><\/div>\n<div id=\"fs-id1167794145876\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167794187159\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167794187159\" class=\"hidden-answer\" style=\"display: none\">\n<ol style=\"list-style-type: lower-alpha;\">\n<li>[latex]\\frac{d}{dx}(\\text{tanh}({x}^{2}+3x))=({\\text{sech}}^{2}({x}^{2}+3x))(2x+3)[\/latex]<\/li>\n<li>[latex]\\frac{d}{dx}(\\frac{1}{{(\\text{sinh}x)}^{2}})=\\frac{d}{dx}{(\\text{sinh}x)}^{-2}=-2{(\\text{sinh}x)}^{-3}\\text{cosh}x[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: Differentiating Hyperbolic Functions and the above Try It.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/ozRCEFk1tmA?controls=0&amp;start=134&amp;end=300&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266834\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266834\" class=\"hidden-answer\" style=\"display: none\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/2.9CalculusOfHyperbolicFunctions134to300_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;2.9 Calculus of Hyperbolic Functions&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<div id=\"fs-id1167793279424\" class=\"textbook exercises\">\n<h3>example: Integrals Involving Hyperbolic Functions<\/h3>\n<p id=\"fs-id1167793965336\">Evaluate the following integrals:<\/p>\n<ol id=\"fs-id1167794050806\" style=\"list-style-type: lower-alpha;\">\n<li>[latex]\\displaystyle\\int x\\text{cosh}({x}^{2})dx[\/latex]<\/li>\n<li>[latex]\\displaystyle\\int \\text{tanh}xdx[\/latex]<\/li>\n<\/ol>\n<div id=\"fs-id1167794051555\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167793254882\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167793254882\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793254882\">We can use [latex]u[\/latex]-substitution in both cases.<\/p>\n<ol id=\"fs-id1167794042760\" style=\"list-style-type: lower-alpha;\">\n<li>Let [latex]u={x}^{2}.[\/latex] Then, [latex]du=2xdx[\/latex] and\n<div id=\"fs-id1167794337482\" class=\"equation unnumbered\">[latex]\\displaystyle\\int x\\text{cosh}({x}^{2})dx=\\displaystyle\\int \\frac{1}{2}\\text{cosh}udu=\\frac{1}{2}\\text{sinh}u+C=\\frac{1}{2}\\text{sinh}({x}^{2})+C.[\/latex]<\/div>\n<\/li>\n<li>Let [latex]u=\\text{cosh}x.[\/latex] Then, [latex]du=\\text{sinh}xdx[\/latex] and\n<div id=\"fs-id1167794043265\" class=\"equation unnumbered\">[latex]\\displaystyle\\int \\text{tanh}xdx=\\displaystyle\\int \\frac{\\text{sinh}x}{\\text{cosh}x}dx=\\displaystyle\\int \\frac{1}{u}du=\\text{ln}|u|+C=\\text{ln}|\\text{cosh}x|+C.[\/latex]<\/div>\n<p>Note that [latex]\\text{cosh}x>0[\/latex] for all [latex]x,[\/latex] so we can eliminate the absolute value signs and obtain<\/p>\n<div id=\"fs-id1167793358384\" class=\"equation unnumbered\">[latex]\\displaystyle\\int \\text{tanh}xdx=\\text{ln}(\\text{cosh}x)+C.[\/latex]<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1167793504384\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1167793819956\">Evaluate the following integrals:<\/p>\n<ol id=\"fs-id1167793985861\" style=\"list-style-type: lower-alpha;\">\n<li>[latex]\\displaystyle\\int {\\text{sinh}}^{3}x\\text{cosh}xdx[\/latex]<\/li>\n<li>[latex]\\displaystyle\\int {\\text{sech}}^{2}(3x)dx[\/latex]<\/li>\n<\/ol>\n<div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q385579\">Hint<\/span><\/p>\n<div id=\"q385579\" class=\"hidden-answer\" style=\"display: none\">Use the formulas above and apply [latex]u[\/latex]-substitution as necessary.<\/div>\n<\/div>\n<\/div>\n<div><\/div>\n<div id=\"fs-id1167794146347\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167793373829\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167793373829\" class=\"hidden-answer\" style=\"display: none\">\n<ol id=\"fs-id1167793373829\" style=\"list-style-type: lower-alpha;\">\n<li>[latex]\\displaystyle\\int {\\text{sinh}}^{3}x\\text{cosh}xdx=\\frac{{\\text{sinh}}^{4}x}{4}+C[\/latex]<\/li>\n<li>[latex]\\displaystyle\\int {\\text{sech}}^{2}(3x)dx=\\frac{\\text{tanh}(3x)}{3}+C[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to the above Try It.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/ozRCEFk1tmA?controls=0&amp;start=478&amp;end=598&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266833\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266833\" class=\"hidden-answer\" style=\"display: none\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/2.9CalculusOfHyperbolicFunctions478to598_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;2.9 Calculus of Hyperbolic Functions&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm223494\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=223494&theme=oea&iframe_resize_id=ohm223494&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Calculus of Inverse Hyperbolic Functions<\/h2>\n<p>Looking at the graphs of the hyperbolic functions, we see that with appropriate range restrictions, they all have inverses. Most of the necessary range restrictions can be discerned by close examination of the graphs. The domains and ranges of the inverse hyperbolic functions are summarized in the following table.<\/p>\n<table id=\"fs-id1167793362111\" summary=\"This table has three columns. The first column is labeled function and has the inverse hyperbolic functions listed in the column. The second column is labeled domain and has the domains of the inverse hyperbolic functions. The third column is labeled range and has the ranges of the inverse hyperbolic functions.\">\n<caption>Domains and Ranges of the Inverse Hyperbolic Functions<\/caption>\n<thead>\n<tr valign=\"top\">\n<th>Function<\/th>\n<th>Domain<\/th>\n<th>Range<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td>[latex]{\\text{sinh}}^{-1}x[\/latex]<\/td>\n<td>[latex](\\text{\u2212}\\infty ,\\infty )[\/latex]<\/td>\n<td>[latex](\\text{\u2212}\\infty ,\\infty )[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]{\\text{cosh}}^{-1}x[\/latex]<\/td>\n<td>[latex](1,\\infty )[\/latex]<\/td>\n<td>[latex][0,\\infty )[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]{\\text{tanh}}^{-1}x[\/latex]<\/td>\n<td>[latex](-1,1)[\/latex]<\/td>\n<td>[latex](\\text{\u2212}\\infty ,\\infty )[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]{\\text{coth}}^{-1}x[\/latex]<\/td>\n<td>[latex](\\text{\u2212}\\infty ,-1)\\cup (1,\\infty )[\/latex]<\/td>\n<td>[latex](\\text{\u2212}\\infty ,0)\\cup (0,\\infty )[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]{\\text{sech}}^{-1}x[\/latex]<\/td>\n<td>[latex](0\\text{, 1})[\/latex]<\/td>\n<td>[latex][0,\\infty )[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]{\\text{csch}}^{-1}x[\/latex]<\/td>\n<td>[latex](\\text{\u2212}\\infty ,0)\\cup (0,\\infty )[\/latex]<\/td>\n<td>[latex](\\text{\u2212}\\infty ,0)\\cup (0,\\infty )[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-id1167793998256\">The graphs of the inverse hyperbolic functions are shown in the following figure.<\/p>\n<div style=\"width: 968px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11213346\/CNX_Calc_Figure_06_09_002.jpg\" alt=\"This figure has six graphs. The first graph labeled \u201ca\u201d is of the function y=sinh^-1(x). It is an increasing function from the 3rd quadrant, through the origin to the first quadrant. The second graph is labeled \u201cb\u201d and is of the function y=cosh^-1(x). It is in the first quadrant, beginning on the x-axis at 2 and increasing. The third graph labeled \u201cc\u201d is of the function y=tanh^-1(x). It is an increasing function from the third quadrant, through the origin, to the first quadrant. The fourth graph is labeled \u201cd\u201d and is of the function y=coth^-1(x). It has two pieces, one in the third quadrant and one in the first quadrant with a vertical asymptote at the y-axis. The fifth graph is labeled \u201ce\u201d and is of the function y=sech^-1(x). It is a curve decreasing in the first quadrant and stopping on the x-axis at x=1. The sixth graph is labeled \u201cf\u201d and is of the function y=csch^-1(x). It has two pieces, one in the third quadrant and one in the first quadrant with a vertical asymptote at the y-axis.\" width=\"958\" height=\"749\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 2. Graphs of the inverse hyperbolic functions.<\/p>\n<\/div>\n<p id=\"fs-id1167793259261\">To find the derivatives of the inverse functions, we use implicit differentiation. We have<\/p>\n<div id=\"fs-id1167793959214\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill y& =\\hfill & {\\text{sinh}}^{-1}x\\hfill \\\\ \\hfill \\text{sinh}y& =\\hfill & x\\hfill \\\\ \\hfill \\frac{d}{dx}\\text{sinh}y& =\\hfill & \\frac{d}{dx}x\\hfill \\\\ \\hfill \\text{cosh}y\\frac{dy}{dx}& =\\hfill & 1.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167793589391\">Recall that [latex]{\\text{cosh}}^{2}y-{\\text{sinh}}^{2}y=1,[\/latex] so [latex]\\text{cosh}y=\\sqrt{1+{\\text{sinh}}^{2}y}.[\/latex] Then,<\/p>\n<div id=\"fs-id1167794042534\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{dy}{dx}=\\frac{1}{\\text{cosh}y}=\\frac{1}{\\sqrt{1+{\\text{sinh}}^{2}y}}=\\frac{1}{\\sqrt{1+{x}^{2}}}.[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167793638871\">We can derive differentiation formulas for the other inverse hyperbolic functions in a similar fashion. These differentiation formulas are summarized in the following table.<\/p>\n<table id=\"fs-id1167794020892\" summary=\"This table has two columns. The first column is labeled f(x) and has the inverse hyperbolic functions as entries. The second column is labeled d\/dx f(x) and is the derivatives of the inverse hyperbolic functions.\">\n<caption>Derivatives of the Inverse Hyperbolic Functions<\/caption>\n<thead>\n<tr valign=\"top\">\n<th>[latex]f(x)[\/latex]<\/th>\n<th>[latex]\\frac{d}{dx}f(x)[\/latex]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td>[latex]{\\text{sinh}}^{-1}x[\/latex]<\/td>\n<td>[latex]\\frac{1}{\\sqrt{1+{x}^{2}}}[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]{\\text{cosh}}^{-1}x[\/latex]<\/td>\n<td>[latex]\\frac{1}{\\sqrt{{x}^{2}-1}}[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]{\\text{tanh}}^{-1}x[\/latex]<\/td>\n<td>[latex]\\frac{1}{1-{x}^{2}}[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]{\\text{coth}}^{-1}x[\/latex]<\/td>\n<td>[latex]\\frac{1}{1-{x}^{2}}[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]{\\text{sech}}^{-1}x[\/latex]<\/td>\n<td>[latex]\\frac{-1}{x\\sqrt{1-{x}^{2}}}[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]{\\text{csch}}^{-1}x[\/latex]<\/td>\n<td>[latex]\\frac{-1}{|x|\\sqrt{1+{x}^{2}}}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-id1167794119173\">Note that the derivatives of [latex]{\\text{tanh}}^{-1}x[\/latex] and [latex]{\\text{coth}}^{-1}x[\/latex] are the same. Thus, when we integrate [latex]1\\text{\/}(1-{x}^{2}),[\/latex] we need to select the proper antiderivative based on the domain of the functions and the values of [latex]x.[\/latex] Integration formulas involving the inverse hyperbolic functions are summarized as follows.<\/p>\n<div id=\"fs-id1167793951927\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cccccccc}\\hfill \\displaystyle\\int \\frac{1}{\\sqrt{1+{u}^{2}}}du& =\\hfill & {\\text{sinh}}^{-1}u+C\\hfill & & & \\hfill \\displaystyle\\int \\frac{1}{u\\sqrt{1-{u}^{2}}}du& =\\hfill & \\text{\u2212}{\\text{sech}}^{-1}|u|+C\\hfill \\\\ \\hfill \\displaystyle\\int \\frac{1}{\\sqrt{{u}^{2}-1}}du& =\\hfill & {\\text{cosh}}^{-1}u+C\\hfill & & & \\hfill \\displaystyle\\int \\frac{1}{u\\sqrt{1+{u}^{2}}}du& =\\hfill & \\text{\u2212}{\\text{csch}}^{-1}|u|+C\\hfill \\\\ \\hfill \\displaystyle\\int \\frac{1}{1-{u}^{2}}du& =\\hfill & \\bigg\\{\\begin{array}{c}{\\text{tanh}}^{-1}u+C\\text{ if }|u|<1\\hfill \\\\ {\\text{coth}}^{-1}u+C\\text{ if }|u|>1\\hfill \\end{array}\\hfill & & & & & \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<div id=\"fs-id1167793984979\" class=\"textbook exercises\">\n<h3>Example: Differentiating Inverse Hyperbolic Functions<\/h3>\n<p id=\"fs-id1167793372563\">Evaluate the following derivatives:<\/p>\n<ol id=\"fs-id1167794155431\" style=\"list-style-type: lower-alpha;\">\n<li>[latex]\\frac{d}{dx}({\\text{sinh}}^{-1}(\\frac{x}{3}))[\/latex]<\/li>\n<li>[latex]\\frac{d}{dx}{({\\text{tanh}}^{-1}x)}^{2}[\/latex]<\/li>\n<\/ol>\n<div id=\"fs-id1167793984981\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167793949787\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167793949787\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793949787\">Using the formulas in the table on derivatives of the inverse hyperbolic functions and the chain rule, we obtain the following results:<\/p>\n<ol id=\"fs-id1167793298499\" style=\"list-style-type: lower-alpha;\">\n<li>[latex]\\frac{d}{dx}({\\text{sinh}}^{-1}(\\frac{x}{3}))=\\frac{1}{3\\sqrt{1+\\frac{{x}^{2}}{9}}}=\\frac{1}{\\sqrt{9+{x}^{2}}}[\/latex]<\/li>\n<li>[latex]\\frac{d}{dx}{({\\text{tanh}}^{-1}x)}^{2}=\\frac{2({\\text{tanh}}^{-1}x)}{1-{x}^{2}}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1167793626577\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1167794332395\">Evaluate the following derivatives:<\/p>\n<ol id=\"fs-id1167793590727\" style=\"list-style-type: lower-alpha;\">\n<li>[latex]\\frac{d}{dx}({\\text{cosh}}^{-1}(3x))[\/latex]<\/li>\n<li>[latex]\\frac{d}{dx}{({\\text{coth}}^{-1}x)}^{3}[\/latex]<\/li>\n<\/ol>\n<div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q358726\">Hint<\/span><\/p>\n<div id=\"q358726\" class=\"hidden-answer\" style=\"display: none\">Use the formulas in the table on derivatives of the inverse hyperbolic functions above and apply the chain rule as necessary.<\/div>\n<\/div>\n<\/div>\n<div><\/div>\n<div id=\"fs-id1167793626580\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167793498614\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167793498614\" class=\"hidden-answer\" style=\"display: none\">\n<ol id=\"fs-id1167793498614\" style=\"list-style-type: lower-alpha;\">\n<li>[latex]\\frac{d}{dx}({\\text{cosh}}^{-1}(3x))=\\frac{3}{\\sqrt{9{x}^{2}-1}}[\/latex]<\/li>\n<li>[latex]\\frac{d}{dx}{({\\text{coth}}^{-1}x)}^{3}=\\frac{3{({\\text{coth}}^{-1}x)}^{2}}{1-{x}^{2}}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to the above Try It.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/ozRCEFk1tmA?controls=0&amp;start=827&amp;end=907&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266836\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266836\" class=\"hidden-answer\" style=\"display: none\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/2.9CalculusOfHyperbolicFunctions827to907_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;2.9 Calculus of Hyperbolic Functions&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<div id=\"fs-id1167793285224\" class=\"textbook exercises\">\n<h3>Example: Integrals Involving Inverse Hyperbolic Functions<\/h3>\n<p id=\"fs-id1167793871641\">Evaluate the following integrals:<\/p>\n<ol id=\"fs-id1167793937415\" style=\"list-style-type: lower-alpha;\">\n<li>[latex]\\displaystyle\\int \\frac{1}{\\sqrt{4{x}^{2}-1}}dx[\/latex]<\/li>\n<li>[latex]\\displaystyle\\int \\frac{1}{2x\\sqrt{1-9{x}^{2}}}dx[\/latex]<\/li>\n<\/ol>\n<div id=\"fs-id1167793285226\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167793361818\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167793361818\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793361818\">We can use [latex]u\\text{-substitution}[\/latex] in both cases.<\/p>\n<ol id=\"fs-id1167794296597\" style=\"list-style-type: lower-alpha;\">\n<li>Let [latex]u=2x.[\/latex] Then, [latex]du=2dx[\/latex] and we have\n<div id=\"fs-id1167793883719\" class=\"equation unnumbered\">[latex]\\displaystyle\\int \\frac{1}{\\sqrt{4{x}^{2}-1}}dx=\\displaystyle\\int \\frac{1}{2\\sqrt{{u}^{2}-1}}du=\\frac{1}{2}{\\text{cosh}}^{-1}u+C=\\frac{1}{2}{\\text{cosh}}^{-1}(2x)+C.[\/latex]<\/div>\n<\/li>\n<li>Let [latex]u=3x.[\/latex] Then, [latex]du=3dx[\/latex] and we obtain\n<div id=\"fs-id1167793444519\" class=\"equation unnumbered\">[latex]\\displaystyle\\int \\frac{1}{2x\\sqrt{1-9{x}^{2}}}dx=\\frac{1}{2}\\displaystyle\\int \\frac{1}{u\\sqrt{1-{u}^{2}}}du=-\\frac{1}{2}{\\text{sech}}^{-1}|u|+C=-\\frac{1}{2}{\\text{sech}}^{-1}|3x|+C.[\/latex]<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1167793925131\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1167793265985\">Evaluate the following integrals:<\/p>\n<ol id=\"fs-id1167793265989\" style=\"list-style-type: lower-alpha;\">\n<li>[latex]\\displaystyle\\int \\frac{1}{\\sqrt{{x}^{2}-4}}dx,\\text{}x>2[\/latex]<\/li>\n<li>[latex]\\displaystyle\\int \\frac{1}{\\sqrt{1-{e}^{2x}}}dx[\/latex]<\/li>\n<\/ol>\n<div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q351257\">Hint<\/span><\/p>\n<div id=\"q351257\" class=\"hidden-answer\" style=\"display: none\">Use the formulas above and apply [latex]u\\text{-substitution}[\/latex] as necessary.<\/div>\n<\/div>\n<\/div>\n<div><\/div>\n<div id=\"fs-id1167793969727\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167793371557\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167793371557\" class=\"hidden-answer\" style=\"display: none\">\n<ol id=\"fs-id1167793371557\" style=\"list-style-type: lower-alpha;\">\n<li>[latex]\\displaystyle\\int \\frac{1}{\\sqrt{{x}^{2}-4}}dx={\\text{cosh}}^{-1}(\\frac{x}{2})+C[\/latex]<\/li>\n<li>[latex]\\displaystyle\\int \\frac{1}{\\sqrt{1-{e}^{2x}}}dx=\\text{\u2212}{\\text{sech}}^{-1}({e}^{x})+C[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to the above Try It.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/ozRCEFk1tmA?controls=0&amp;start=1140&amp;end=1263&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266837\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266837\" class=\"hidden-answer\" style=\"display: none\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/2.9CalculusOfHyperbolicFunctions1140to1263_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;2.9 Calculus of Hyperbolic Functions&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm223496\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=223496&theme=oea&iframe_resize_id=ohm223496&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-1195\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>2.9 Calculus of Hyperbolic Functions. <strong>Authored by<\/strong>: Ryan Melton. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Calculus Volume 2. <strong>Authored by<\/strong>: Gilbert Strang, Edwin (Jed) Herman. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction\">https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":17533,"menu_order":35,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 2\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"2.9 Calculus of Hyperbolic Functions\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1195","chapter","type-chapter","status-publish","hentry"],"part":1160,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1195","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/users\/17533"}],"version-history":[{"count":2,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1195\/revisions"}],"predecessor-version":[{"id":1406,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1195\/revisions\/1406"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/1160"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1195\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=1195"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=1195"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=1195"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=1195"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}