{"id":1140,"date":"2021-06-30T17:01:59","date_gmt":"2021-06-30T17:01:59","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/integrals-involving-logarithmic-functions\/"},"modified":"2022-03-19T03:22:35","modified_gmt":"2022-03-19T03:22:35","slug":"integrals-involving-logarithmic-functions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/integrals-involving-logarithmic-functions\/","title":{"raw":"Integrals Involving Logarithmic Functions","rendered":"Integrals Involving Logarithmic Functions"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Integrate functions involving logarithmic functions<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-id1170572565333\">Integrating functions of the form [latex]f(x)={x}^{-1}[\/latex] result in the absolute value of the natural log function, as shown in the following rule. Integral formulas for other logarithmic functions, such as [latex]f(x)=\\text{ln}x[\/latex] and [latex]f(x)={\\text{log}}_{a}x,[\/latex] are also included in the rule.<\/p>\r\n\r\n<div id=\"fs-id1170572543658\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Integration Formulas Involving Logarithmic Functions<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1170572543663\">The following formulas can be used to evaluate integrals involving logarithmic functions.<\/p>\r\n\r\n<div id=\"fs-id1170572543666\" class=\"equation\" style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill \\displaystyle\\int {x}^{-1}dx&amp; =\\hfill &amp; \\text{ln}|x|+C\\hfill \\\\ \\hfill \\displaystyle\\int \\text{ln}xdx&amp; =\\hfill &amp; x\\text{ln}x-x+C=x(\\text{ln}x-1)+C\\hfill \\\\ \\hfill \\displaystyle\\int {\\text{log}}_{a}xdx&amp; =\\hfill &amp; \\frac{x}{\\text{ln}a}(\\text{ln}x-1)+C\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n<div id=\"fs-id1170571653392\" class=\"textbook exercises\">\r\n<h3>Example: Finding an Antiderivative Involving [latex]\\text{ln}x[\/latex]<\/h3>\r\nFind the antiderivative of the function [latex]\\dfrac{3}{x-10}.[\/latex]\r\n<div id=\"fs-id1170571653394\" class=\"exercise\">[reveal-answer q=\"fs-id1170571653435\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571653435\"]\r\n<p id=\"fs-id1170571653435\">First factor the 3 outside the integral symbol. Then use the [latex]u^{-1}[\/latex]\u00a0rule. Thus,<\/p>\r\n\r\n<div id=\"fs-id1170571653444\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll}{\\displaystyle\\int \\frac{3}{x-10}dx}\\hfill &amp; =3{\\displaystyle\\int \\frac{1}{x-10}dx}\\hfill \\\\ \\\\ \\\\ &amp; =3{\\displaystyle\\int \\frac{du}{u}}\\hfill \\\\ &amp; =3\\text{ln}|u|+C\\hfill \\\\ &amp; =3\\text{ln}|x-10|+C,x\\ne 10.\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1170571597433\">See Figure 3.<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"325\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11204256\/CNX_Calc_Figure_05_06_003.jpg\" alt=\"A graph of the function f(x) = 3 \/ (x \u2013 10). There is an asymptote at x=10. The first segment is a decreasing concave down curve that approaches 0 as x goes to negative infinity and approaches negative infinity as x goes to 10. The second segment is a decreasing concave up curve that approaches infinity as x goes to 10 and approaches 0 as x approaches infinity.\" width=\"325\" height=\"246\" \/> Figure 3. The domain of this function is [latex]x\\ne 10.[\/latex][\/caption][\/hidden-answer]<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571597473\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nFind the antiderivative of [latex]\\dfrac{1}{x+2}.[\/latex]\r\n\r\n[reveal-answer q=\"184928\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"184928\"]Follow the pattern from the last example to solve the problem.[\/hidden-answer]\r\n<div id=\"fs-id1170571597476\" class=\"exercise\">[reveal-answer q=\"fs-id1170572480516\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572480516\"]\r\n<p id=\"fs-id1170572480516\">[latex]\\text{ln}|x+2|+C[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170572480542\" class=\"textbook exercises\">\r\n<h3>Example: Finding an Antiderivative of a Rational Function<\/h3>\r\nFind the antiderivative of [latex]\\dfrac{2{x}^{3}+3x}{{x}^{4}+3{x}^{2}}.[\/latex]\r\n<div id=\"fs-id1170572480545\" class=\"exercise\">\r\n<div class=\"solution\">[reveal-answer q=\"fs-id1170572480517\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572480517\"]\r\nThis can be rewritten as [latex]\\displaystyle\\int (2{x}^{3}+3x){({x}^{4}+3{x}^{2})}^{-1}dx.[\/latex] Use substitution. Let [latex]u={x}^{4}+3{x}^{2},[\/latex] then [latex]du=4{x}^{3}+6x.[\/latex] Alter <em>du<\/em> by factoring out the 2. Thus,\r\n<div id=\"fs-id1170571807881\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{}\\\\ \\hfill du&amp; =\\hfill &amp; (4{x}^{3}+6x)dx\\hfill \\\\ &amp; =\\hfill &amp; 2(2{x}^{3}+3x)dx\\hfill \\\\ \\hfill \\frac{1}{2}du&amp; =\\hfill &amp; (2{x}^{3}+3x)dx.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170572373464\">Rewrite the integrand in [latex]u[\/latex]:<\/p>\r\n\r\n<div id=\"fs-id1170572331833\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\int (2{x}^{3}+3x){({x}^{4}+3{x}^{2})}^{-1}dx=\\frac{1}{2}\\displaystyle\\int {u}^{-1}du.[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170571733965\">Then we have<\/p>\r\n\r\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll}\\frac{1}{2}{\\displaystyle\\int {u}^{-1}du}\\hfill &amp; =\\frac{1}{2}\\text{ln}|u|+C\\hfill \\\\ \\\\ &amp; =\\frac{1}{2}\\text{ln}|{x}^{4}+3{x}^{2}|+C.\\hfill \\end{array}[\/latex]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\nWatch the following video to see the worked solution to Example: Finding an Antiderivative of a Rational Function.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/8mW2bG6HoPE?controls=0&amp;start=162&amp;end=270&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\/5.6.2_162to270_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"5.6.2\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<div id=\"fs-id1170571649922\" class=\"textbook exercises\">\r\n<div class=\"exercise\">\r\n<h3>Example: Finding an Antiderivative of a Logarithmic Function<\/h3>\r\nFind the antiderivative of the log function [latex]{\\text{log}}_{2}x.[\/latex]\r\n[reveal-answer q=\"fs-id1170571649954\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571649954\"]\r\n<p id=\"fs-id1170571649954\">Follow the format in the formula listed in the rule on integration formulas involving logarithmic functions. Based on this format, we have<\/p>\r\n\r\n<div id=\"fs-id1170571649959\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\int {\\text{log}}_{2}xdx=\\frac{x}{\\text{ln}2}(\\text{ln}x-1)+C.[\/latex][\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571660181\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nFind the antiderivative of [latex]{\\text{log}}_{3}x.[\/latex]\r\n\r\n[reveal-answer q=\"697008\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"697008\"]Follow the previous example and refer to the rule on integration formulas involving logarithmic functions.[\/hidden-answer]\r\n<div id=\"fs-id1170571660184\" class=\"exercise\">[reveal-answer q=\"fs-id1170571660214\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571660214\"]\r\n<p id=\"fs-id1170571660214\">[latex]\\frac{x}{\\text{ln}3}(\\text{ln}x-1)+C[\/latex]<\/p>\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\/8mW2bG6HoPE?controls=0&amp;start=274&amp;end=342&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\/5.6.2_274to342_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"5.6.2\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n\r\nThe example below is a definite integral of a trigonometric function. With trigonometric functions, we often have to apply a trigonometric property or an identity before we can move forward. Finding the right form of the integrand is usually the key to a smooth integration.\r\n<div class=\"textbox exercises\">\r\n<h3>Evaluating a Definite Integral<\/h3>\r\nFind the definite integral of [latex]{\\displaystyle\\int }_{0}^{\\pi \\text{\/}2}\\frac{ \\sin x}{1+ \\cos x}dx.[\/latex]\r\n<div id=\"fs-id1170572274910\" class=\"exercise\">[reveal-answer q=\"fs-id1170571636148\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571636148\"]\r\n<p id=\"fs-id1170571636148\">We need substitution to evaluate this problem. Let [latex]u=1+ \\cos x,,[\/latex] so [latex]du=\\text{\u2212} \\sin xdx.[\/latex] Rewrite the integral in terms of [latex]u[\/latex], changing the limits of integration as well. Thus,<\/p>\r\n\r\n<div id=\"fs-id1170571636203\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{c}u=1+ \\cos (0)=2\\hfill \\\\ u=1+ \\cos (\\frac{\\pi }{2})=1\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170572510108\">Then<\/p>\r\n\r\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}{\\displaystyle\\int} _{0}^{\\pi \\text{\/}2}\\dfrac{ \\sin x}{1+ \\cos x}\\hfill &amp; =\\text{\u2212}{\\displaystyle\\int }_{2}^{1}{u}^{-1}du\\hfill \\\\ \\\\ \\\\ &amp; ={\\displaystyle\\int }_{1}^{2}{u}^{-1}du\\hfill \\\\ &amp; ={\\text{ln}|u||}_{1}^{2}\\hfill \\\\ &amp; =\\left[\\text{ln}2-\\text{ln}1\\right]\\hfill \\\\ &amp; =\\text{ln}2\\hfill \\end{array}[\/latex][\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]20050[\/ohm_question]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Integrate functions involving logarithmic functions<\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-id1170572565333\">Integrating functions of the form [latex]f(x)={x}^{-1}[\/latex] result in the absolute value of the natural log function, as shown in the following rule. Integral formulas for other logarithmic functions, such as [latex]f(x)=\\text{ln}x[\/latex] and [latex]f(x)={\\text{log}}_{a}x,[\/latex] are also included in the rule.<\/p>\n<div id=\"fs-id1170572543658\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Integration Formulas Involving Logarithmic Functions<\/h3>\n<hr \/>\n<p id=\"fs-id1170572543663\">The following formulas can be used to evaluate integrals involving logarithmic functions.<\/p>\n<div id=\"fs-id1170572543666\" class=\"equation\" style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill \\displaystyle\\int {x}^{-1}dx& =\\hfill & \\text{ln}|x|+C\\hfill \\\\ \\hfill \\displaystyle\\int \\text{ln}xdx& =\\hfill & x\\text{ln}x-x+C=x(\\text{ln}x-1)+C\\hfill \\\\ \\hfill \\displaystyle\\int {\\text{log}}_{a}xdx& =\\hfill & \\frac{x}{\\text{ln}a}(\\text{ln}x-1)+C\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<div id=\"fs-id1170571653392\" class=\"textbook exercises\">\n<h3>Example: Finding an Antiderivative Involving [latex]\\text{ln}x[\/latex]<\/h3>\n<p>Find the antiderivative of the function [latex]\\dfrac{3}{x-10}.[\/latex]<\/p>\n<div id=\"fs-id1170571653394\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170571653435\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170571653435\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571653435\">First factor the 3 outside the integral symbol. Then use the [latex]u^{-1}[\/latex]\u00a0rule. Thus,<\/p>\n<div id=\"fs-id1170571653444\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll}{\\displaystyle\\int \\frac{3}{x-10}dx}\\hfill & =3{\\displaystyle\\int \\frac{1}{x-10}dx}\\hfill \\\\ \\\\ \\\\ & =3{\\displaystyle\\int \\frac{du}{u}}\\hfill \\\\ & =3\\text{ln}|u|+C\\hfill \\\\ & =3\\text{ln}|x-10|+C,x\\ne 10.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1170571597433\">See Figure 3.<\/p>\n<div style=\"width: 335px\" 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\/11204256\/CNX_Calc_Figure_05_06_003.jpg\" alt=\"A graph of the function f(x) = 3 \/ (x \u2013 10). There is an asymptote at x=10. The first segment is a decreasing concave down curve that approaches 0 as x goes to negative infinity and approaches negative infinity as x goes to 10. The second segment is a decreasing concave up curve that approaches infinity as x goes to 10 and approaches 0 as x approaches infinity.\" width=\"325\" height=\"246\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 3. The domain of this function is [latex]x\\ne 10.[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571597473\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>Find the antiderivative of [latex]\\dfrac{1}{x+2}.[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q184928\">Hint<\/span><\/p>\n<div id=\"q184928\" class=\"hidden-answer\" style=\"display: none\">Follow the pattern from the last example to solve the problem.<\/div>\n<\/div>\n<div id=\"fs-id1170571597476\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572480516\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572480516\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572480516\">[latex]\\text{ln}|x+2|+C[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170572480542\" class=\"textbook exercises\">\n<h3>Example: Finding an Antiderivative of a Rational Function<\/h3>\n<p>Find the antiderivative of [latex]\\dfrac{2{x}^{3}+3x}{{x}^{4}+3{x}^{2}}.[\/latex]<\/p>\n<div id=\"fs-id1170572480545\" class=\"exercise\">\n<div class=\"solution\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572480517\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572480517\" class=\"hidden-answer\" style=\"display: none\">\nThis can be rewritten as [latex]\\displaystyle\\int (2{x}^{3}+3x){({x}^{4}+3{x}^{2})}^{-1}dx.[\/latex] Use substitution. Let [latex]u={x}^{4}+3{x}^{2},[\/latex] then [latex]du=4{x}^{3}+6x.[\/latex] Alter <em>du<\/em> by factoring out the 2. Thus,<\/p>\n<div id=\"fs-id1170571807881\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{}\\\\ \\hfill du& =\\hfill & (4{x}^{3}+6x)dx\\hfill \\\\ & =\\hfill & 2(2{x}^{3}+3x)dx\\hfill \\\\ \\hfill \\frac{1}{2}du& =\\hfill & (2{x}^{3}+3x)dx.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572373464\">Rewrite the integrand in [latex]u[\/latex]:<\/p>\n<div id=\"fs-id1170572331833\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\int (2{x}^{3}+3x){({x}^{4}+3{x}^{2})}^{-1}dx=\\frac{1}{2}\\displaystyle\\int {u}^{-1}du.[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170571733965\">Then we have<\/p>\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll}\\frac{1}{2}{\\displaystyle\\int {u}^{-1}du}\\hfill & =\\frac{1}{2}\\text{ln}|u|+C\\hfill \\\\ \\\\ & =\\frac{1}{2}\\text{ln}|{x}^{4}+3{x}^{2}|+C.\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: Finding an Antiderivative of a Rational Function.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/8mW2bG6HoPE?controls=0&amp;start=162&amp;end=270&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\/5.6.2_162to270_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;5.6.2&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<div id=\"fs-id1170571649922\" class=\"textbook exercises\">\n<div class=\"exercise\">\n<h3>Example: Finding an Antiderivative of a Logarithmic Function<\/h3>\n<p>Find the antiderivative of the log function [latex]{\\text{log}}_{2}x.[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170571649954\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170571649954\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571649954\">Follow the format in the formula listed in the rule on integration formulas involving logarithmic functions. Based on this format, we have<\/p>\n<div id=\"fs-id1170571649959\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\int {\\text{log}}_{2}xdx=\\frac{x}{\\text{ln}2}(\\text{ln}x-1)+C.[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571660181\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>Find the antiderivative of [latex]{\\text{log}}_{3}x.[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q697008\">Hint<\/span><\/p>\n<div id=\"q697008\" class=\"hidden-answer\" style=\"display: none\">Follow the previous example and refer to the rule on integration formulas involving logarithmic functions.<\/div>\n<\/div>\n<div id=\"fs-id1170571660184\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170571660214\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170571660214\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571660214\">[latex]\\frac{x}{\\text{ln}3}(\\text{ln}x-1)+C[\/latex]<\/p>\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\/8mW2bG6HoPE?controls=0&amp;start=274&amp;end=342&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\/5.6.2_274to342_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;5.6.2&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<p>The example below is a definite integral of a trigonometric function. With trigonometric functions, we often have to apply a trigonometric property or an identity before we can move forward. Finding the right form of the integrand is usually the key to a smooth integration.<\/p>\n<div class=\"textbox exercises\">\n<h3>Evaluating a Definite Integral<\/h3>\n<p>Find the definite integral of [latex]{\\displaystyle\\int }_{0}^{\\pi \\text{\/}2}\\frac{ \\sin x}{1+ \\cos x}dx.[\/latex]<\/p>\n<div id=\"fs-id1170572274910\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170571636148\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170571636148\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571636148\">We need substitution to evaluate this problem. Let [latex]u=1+ \\cos x,,[\/latex] so [latex]du=\\text{\u2212} \\sin xdx.[\/latex] Rewrite the integral in terms of [latex]u[\/latex], changing the limits of integration as well. Thus,<\/p>\n<div id=\"fs-id1170571636203\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{c}u=1+ \\cos (0)=2\\hfill \\\\ u=1+ \\cos (\\frac{\\pi }{2})=1\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572510108\">Then<\/p>\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}{\\displaystyle\\int} _{0}^{\\pi \\text{\/}2}\\dfrac{ \\sin x}{1+ \\cos x}\\hfill & =\\text{\u2212}{\\displaystyle\\int }_{2}^{1}{u}^{-1}du\\hfill \\\\ \\\\ \\\\ & ={\\displaystyle\\int }_{1}^{2}{u}^{-1}du\\hfill \\\\ & ={\\text{ln}|u||}_{1}^{2}\\hfill \\\\ & =\\left[\\text{ln}2-\\text{ln}1\\right]\\hfill \\\\ & =\\text{ln}2\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm20050\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=20050&theme=oea&iframe_resize_id=ohm20050&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-1140\">\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>5.6.2. <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":27,"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\":\"5.6.2\",\"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-1140","chapter","type-chapter","status-publish","hentry"],"part":1113,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1140","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\/1140\/revisions"}],"predecessor-version":[{"id":1370,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1140\/revisions\/1370"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/1113"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1140\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=1140"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=1140"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=1140"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=1140"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}