{"id":1974,"date":"2021-08-19T16:03:43","date_gmt":"2021-08-19T16:03:43","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/review-for-integrals-resulting-in-inverse-trigonometric-functions\/"},"modified":"2021-11-17T01:32:46","modified_gmt":"2021-11-17T01:32:46","slug":"review-for-integrals-resulting-in-inverse-trigonometric-functions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/review-for-integrals-resulting-in-inverse-trigonometric-functions\/","title":{"raw":"Skills Review for Integrals Resulting in Inverse Trigonometric Functions","rendered":"Skills Review for Integrals Resulting in Inverse Trigonometric Functions"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Evaluate inverse trigonometric functions<\/li>\r\n<\/ul>\r\n<\/div>\r\nIn the Integrals Resulting in Inverse Trigonometric Functions section, we will need to know how to evaluate inverse trigonometric functions. This concept is reviewed here.\r\n<h2>Evaluate Inverse Trigonometric Functions<\/h2>\r\nIn order to use inverse trigonometric functions, we need to understand that an inverse trigonometric function \u201cundoes\u201d what the original trigonometric function \u201cdoes,\u201d as is the case with any other function and its inverse. In other words, the domain of the inverse function is the range of the original function, and vice versa, as summarized below.\r\n\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27163959\/CNX_Precalc_Figure_06_03_013.jpg\" alt=\"A chart that says \u201cTrig Functinos\u201d, \u201cInverse Trig Functions\u201d, \u201cDomain: Measure of an angle\u201d, \u201cDomain: Ratio\u201d, \u201cRange: Ratio\u201d, and \u201cRange: Measure of an angle\u201d.\" width=\"731\" height=\"78\" \/>\r\n\r\nFor example, if [latex]f(x)=\\sin x[\/latex], then we would write\u00a0[latex]f^{-1}(x)={\\sin}^{-1}{x}[\/latex]. Be aware that [latex]{\\sin}^{-1}x[\/latex] does not mean [latex]\\frac{1}{\\sin{x}}[\/latex]. The following examples illustrate the inverse trigonometric functions:\r\n<ul>\r\n \t<li>Since [latex]\\sin\\left(\\frac{\\pi}{6}\\right)=\\frac{1}{2}[\/latex], then [latex]\\frac{\\pi}{6}=\\sin^{\u22121}(\\frac{1}{2})[\/latex].<\/li>\r\n \t<li>Since [latex]\\cos(\\pi)=\u22121[\/latex], then [latex]\\pi=\\cos^{\u22121}(\u22121)[\/latex].<\/li>\r\n \t<li>Since [latex]\\tan\\left(\\frac{\\pi}{4}\\right)=1[\/latex], then [latex]\\frac{\\pi}{4}=\\tan^{\u22121}(1)[\/latex].<\/li>\r\n<\/ul>\r\n<div>\r\n<div style=\"text-align: center;\"><\/div>\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>A General Note: Relations for Inverse Sine, Cosine, and Tangent Functions<\/h3>\r\nFor angles in the interval [latex]\\left[\u2212\\frac{\\pi}{2}\\text{, }\\frac{\\pi}{2}\\right][\/latex], if [latex]\\sin y=x[\/latex], then [latex]\\sin^{\u22121}x=y[\/latex].\r\n\r\nFor angles in the interval [0, \u03c0], if [latex]\\cos y=x[\/latex], then [latex]\\cos^{\u22121}x=y[\/latex].\r\n\r\nFor angles in the interval [latex]\\left(\u2212\\frac{\\pi}{2}\\text{, }\\frac{\\pi}{2}\\right)[\/latex], if [latex]\\tan y=x[\/latex], then [latex]\\tan^{\u22121}x=y[\/latex].\r\n\r\n<\/div>\r\nJust as we did with the original trigonometric functions, we can give exact values for the inverse functions when we are using the special angles, specifically [latex]\\frac{\\pi}{ 6} (30^\\circ)\\text{, }\\frac{\\pi}{ 4} (45^\\circ),\\text{ and } \\frac{\\pi}{ 3} (60^\\circ)[\/latex], and their reflections into other quadrants.\r\n<div class=\"textbox\">\r\n<h3>How To: Given a \u201cspecial\u201d input value, evaluate an inverse trigonometric function.<\/h3>\r\n<ol>\r\n \t<li>Find angle\u00a0<em>x<\/em>\u00a0for which the original trigonometric function has an output equal to the given input for the inverse trigonometric function.<\/li>\r\n \t<li>If\u00a0<em>x<\/em>\u00a0is not in the defined range of the inverse, find another angle\u00a0<em>y<\/em>\u00a0that is in the defined range and has the same sine, cosine, or tangent as\u00a0<em>x<\/em>, depending on which corresponds to the given inverse function.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Evaluating Inverse Trigonometric Functions for Special Input Values<\/h3>\r\nEvaluate each of the following.\r\n<p style=\"padding-left: 60px;\">a. [latex]\\sin\u22121\\left(\\frac{1}{2}\\right)[\/latex]<\/p>\r\n<p style=\"padding-left: 60px;\">b. [latex]\\sin\u22121\\left(\u2212\\frac{2}{\\sqrt{2}}\\right)[\/latex]<\/p>\r\n<p style=\"padding-left: 60px;\">c. [latex]\\cos\u22121\\left(\u2212\\frac{3}{\\sqrt{2}}\\right)[\/latex]<\/p>\r\n<p style=\"padding-left: 60px;\">d. [latex]\\tan^{\u2212 1}(1)[\/latex]<\/p>\r\n[reveal-answer q=\"666370\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"666370\"]\r\n<p style=\"padding-left: 60px;\">a. Evaluating [latex]\\sin^{\u22121}(\\frac{1}{2})[\/latex] is the same as determining the angle that would have a sine value of [latex]\\frac{1}{2}[\/latex]. In other words, what angle <em>x<\/em> would satisfy [latex]\\sin(x)=\\frac{1}{2}[\/latex]? There are multiple values that would satisfy this relationship, such as [latex]\\frac{\\pi}{6}[\/latex] and [latex]\\frac{5\\pi}{6}[\/latex], but we know we need the angle in the interval [latex]\\left[\u2212\\frac{\\pi}{2}\\text{, }\\frac{\\pi}{2}\\right][\/latex], so the answer will be [latex]\\sin^{\u22121}(\\frac{1}{2})=\\frac{\\pi}{6}[\/latex]. Remember that the inverse is a function, so for each input, we will get exactly one output.<\/p>\r\n<p style=\"padding-left: 60px;\">b. To evaluate [latex]\\sin^{\u22121}\\left(\u2212\\frac{\\sqrt{2}}{2}\\right)[\/latex], we know that [latex]\\frac{5\\pi}{4}[\/latex] and [latex]\\frac{7\\pi}{4}[\/latex] both have a sine value of [latex]\u2212\\frac{\\sqrt{2}}{2}[\/latex], but neither is in the interval [latex]\\left[\u2212\\frac{\\pi}{2}\\text{, }\\frac{\\pi}{2}\\right][\/latex]. For that, we need the negative angle coterminal with [latex]\\frac{7\\pi}{4}:\\sin^{\u22121}\\left(\u2212\\frac{\\sqrt{2}}{2}\\right)=\u2212\\frac{\\pi}{4}[\/latex].<\/p>\r\n<p style=\"padding-left: 60px;\">c. To evaluate [latex]\\cos^{\u22121}\\left(\u2212\\frac{\\sqrt{3}}{2}\\right)[\/latex], we are looking for an angle in the interval [0,\u03c0] with a cosine value of [latex]\u2212\\frac{\\sqrt{3}}{2}[\/latex]. The angle that satisfies this is [latex]\\cos^{\u22121}\\left(\u2212\\frac{\\sqrt{3}}{2}\\right)=\\frac{5\\pi}{6}[\/latex].<\/p>\r\n<p style=\"padding-left: 60px;\">d. Evaluating [latex]\\tan^{\u22121}(1)[\/latex], we are looking for an angle in the interval [latex](\u2212\\frac{\\pi}{2}\\text{, }\\frac{\\pi}{2})[\/latex] with a tangent value of 1. The correct angle is [latex]\\tan^{\u22121}(1)=\\frac{\\pi}{4}[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nEvaluate each of the following.\r\n<ol>\r\n \t<li>[latex]\\sin^{\u22121}(\u22121)[\/latex]<\/li>\r\n \t<li>[latex]\\tan^{\u22121}(\u22121)[\/latex]<\/li>\r\n \t<li>[latex]\\cos^{\u22121}(\u22121)[\/latex]<\/li>\r\n \t<li>[latex]\\cos^{\u22121}(\\frac{1}{2})[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"333778\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"333778\"]\r\n\r\n1. [latex]\u2212\\frac{\\pi}{2}[\/latex];\r\n\r\n2. [latex]\u2212\\frac{\\pi}{4}[\/latex]\r\n\r\n3. [latex]\\pi[\/latex]\r\n\r\n4. [latex]\\frac{\\pi}{3}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]173433[\/ohm_question]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Evaluate inverse trigonometric functions<\/li>\n<\/ul>\n<\/div>\n<p>In the Integrals Resulting in Inverse Trigonometric Functions section, we will need to know how to evaluate inverse trigonometric functions. This concept is reviewed here.<\/p>\n<h2>Evaluate Inverse Trigonometric Functions<\/h2>\n<p>In order to use inverse trigonometric functions, we need to understand that an inverse trigonometric function \u201cundoes\u201d what the original trigonometric function \u201cdoes,\u201d as is the case with any other function and its inverse. In other words, the domain of the inverse function is the range of the original function, and vice versa, as summarized below.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27163959\/CNX_Precalc_Figure_06_03_013.jpg\" alt=\"A chart that says \u201cTrig Functinos\u201d, \u201cInverse Trig Functions\u201d, \u201cDomain: Measure of an angle\u201d, \u201cDomain: Ratio\u201d, \u201cRange: Ratio\u201d, and \u201cRange: Measure of an angle\u201d.\" width=\"731\" height=\"78\" \/><\/p>\n<p>For example, if [latex]f(x)=\\sin x[\/latex], then we would write\u00a0[latex]f^{-1}(x)={\\sin}^{-1}{x}[\/latex]. Be aware that [latex]{\\sin}^{-1}x[\/latex] does not mean [latex]\\frac{1}{\\sin{x}}[\/latex]. The following examples illustrate the inverse trigonometric functions:<\/p>\n<ul>\n<li>Since [latex]\\sin\\left(\\frac{\\pi}{6}\\right)=\\frac{1}{2}[\/latex], then [latex]\\frac{\\pi}{6}=\\sin^{\u22121}(\\frac{1}{2})[\/latex].<\/li>\n<li>Since [latex]\\cos(\\pi)=\u22121[\/latex], then [latex]\\pi=\\cos^{\u22121}(\u22121)[\/latex].<\/li>\n<li>Since [latex]\\tan\\left(\\frac{\\pi}{4}\\right)=1[\/latex], then [latex]\\frac{\\pi}{4}=\\tan^{\u22121}(1)[\/latex].<\/li>\n<\/ul>\n<div>\n<div style=\"text-align: center;\"><\/div>\n<\/div>\n<div class=\"textbox\">\n<h3>A General Note: Relations for Inverse Sine, Cosine, and Tangent Functions<\/h3>\n<p>For angles in the interval [latex]\\left[\u2212\\frac{\\pi}{2}\\text{, }\\frac{\\pi}{2}\\right][\/latex], if [latex]\\sin y=x[\/latex], then [latex]\\sin^{\u22121}x=y[\/latex].<\/p>\n<p>For angles in the interval [0, \u03c0], if [latex]\\cos y=x[\/latex], then [latex]\\cos^{\u22121}x=y[\/latex].<\/p>\n<p>For angles in the interval [latex]\\left(\u2212\\frac{\\pi}{2}\\text{, }\\frac{\\pi}{2}\\right)[\/latex], if [latex]\\tan y=x[\/latex], then [latex]\\tan^{\u22121}x=y[\/latex].<\/p>\n<\/div>\n<p>Just as we did with the original trigonometric functions, we can give exact values for the inverse functions when we are using the special angles, specifically [latex]\\frac{\\pi}{ 6} (30^\\circ)\\text{, }\\frac{\\pi}{ 4} (45^\\circ),\\text{ and } \\frac{\\pi}{ 3} (60^\\circ)[\/latex], and their reflections into other quadrants.<\/p>\n<div class=\"textbox\">\n<h3>How To: Given a \u201cspecial\u201d input value, evaluate an inverse trigonometric function.<\/h3>\n<ol>\n<li>Find angle\u00a0<em>x<\/em>\u00a0for which the original trigonometric function has an output equal to the given input for the inverse trigonometric function.<\/li>\n<li>If\u00a0<em>x<\/em>\u00a0is not in the defined range of the inverse, find another angle\u00a0<em>y<\/em>\u00a0that is in the defined range and has the same sine, cosine, or tangent as\u00a0<em>x<\/em>, depending on which corresponds to the given inverse function.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Evaluating Inverse Trigonometric Functions for Special Input Values<\/h3>\n<p>Evaluate each of the following.<\/p>\n<p style=\"padding-left: 60px;\">a. [latex]\\sin\u22121\\left(\\frac{1}{2}\\right)[\/latex]<\/p>\n<p style=\"padding-left: 60px;\">b. [latex]\\sin\u22121\\left(\u2212\\frac{2}{\\sqrt{2}}\\right)[\/latex]<\/p>\n<p style=\"padding-left: 60px;\">c. [latex]\\cos\u22121\\left(\u2212\\frac{3}{\\sqrt{2}}\\right)[\/latex]<\/p>\n<p style=\"padding-left: 60px;\">d. [latex]\\tan^{\u2212 1}(1)[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q666370\">Show Solution<\/span><\/p>\n<div id=\"q666370\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"padding-left: 60px;\">a. Evaluating [latex]\\sin^{\u22121}(\\frac{1}{2})[\/latex] is the same as determining the angle that would have a sine value of [latex]\\frac{1}{2}[\/latex]. In other words, what angle <em>x<\/em> would satisfy [latex]\\sin(x)=\\frac{1}{2}[\/latex]? There are multiple values that would satisfy this relationship, such as [latex]\\frac{\\pi}{6}[\/latex] and [latex]\\frac{5\\pi}{6}[\/latex], but we know we need the angle in the interval [latex]\\left[\u2212\\frac{\\pi}{2}\\text{, }\\frac{\\pi}{2}\\right][\/latex], so the answer will be [latex]\\sin^{\u22121}(\\frac{1}{2})=\\frac{\\pi}{6}[\/latex]. Remember that the inverse is a function, so for each input, we will get exactly one output.<\/p>\n<p style=\"padding-left: 60px;\">b. To evaluate [latex]\\sin^{\u22121}\\left(\u2212\\frac{\\sqrt{2}}{2}\\right)[\/latex], we know that [latex]\\frac{5\\pi}{4}[\/latex] and [latex]\\frac{7\\pi}{4}[\/latex] both have a sine value of [latex]\u2212\\frac{\\sqrt{2}}{2}[\/latex], but neither is in the interval [latex]\\left[\u2212\\frac{\\pi}{2}\\text{, }\\frac{\\pi}{2}\\right][\/latex]. For that, we need the negative angle coterminal with [latex]\\frac{7\\pi}{4}:\\sin^{\u22121}\\left(\u2212\\frac{\\sqrt{2}}{2}\\right)=\u2212\\frac{\\pi}{4}[\/latex].<\/p>\n<p style=\"padding-left: 60px;\">c. To evaluate [latex]\\cos^{\u22121}\\left(\u2212\\frac{\\sqrt{3}}{2}\\right)[\/latex], we are looking for an angle in the interval [0,\u03c0] with a cosine value of [latex]\u2212\\frac{\\sqrt{3}}{2}[\/latex]. The angle that satisfies this is [latex]\\cos^{\u22121}\\left(\u2212\\frac{\\sqrt{3}}{2}\\right)=\\frac{5\\pi}{6}[\/latex].<\/p>\n<p style=\"padding-left: 60px;\">d. Evaluating [latex]\\tan^{\u22121}(1)[\/latex], we are looking for an angle in the interval [latex](\u2212\\frac{\\pi}{2}\\text{, }\\frac{\\pi}{2})[\/latex] with a tangent value of 1. The correct angle is [latex]\\tan^{\u22121}(1)=\\frac{\\pi}{4}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Evaluate each of the following.<\/p>\n<ol>\n<li>[latex]\\sin^{\u22121}(\u22121)[\/latex]<\/li>\n<li>[latex]\\tan^{\u22121}(\u22121)[\/latex]<\/li>\n<li>[latex]\\cos^{\u22121}(\u22121)[\/latex]<\/li>\n<li>[latex]\\cos^{\u22121}(\\frac{1}{2})[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q333778\">Show Solution<\/span><\/p>\n<div id=\"q333778\" class=\"hidden-answer\" style=\"display: none\">\n<p>1. [latex]\u2212\\frac{\\pi}{2}[\/latex];<\/p>\n<p>2. [latex]\u2212\\frac{\\pi}{4}[\/latex]<\/p>\n<p>3. [latex]\\pi[\/latex]<\/p>\n<p>4. [latex]\\frac{\\pi}{3}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm173433\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=173433&theme=oea&iframe_resize_id=ohm173433\" 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-1974\">\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>Modification and Revision . <strong>Provided by<\/strong>: Lumen Learning. <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>College Algebra Corequisite. <strong>Provided by<\/strong>: Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/courses.lumenlearning.com\/waymakercollegealgebracorequisite\/\">https:\/\/courses.lumenlearning.com\/waymakercollegealgebracorequisite\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Precalculus. <strong>Provided by<\/strong>: Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/courses.lumenlearning.com\/precalculus\/\">https:\/\/courses.lumenlearning.com\/precalculus\/<\/a>. <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>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":349141,"menu_order":6,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"Modification and Revision \",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"College Algebra Corequisite\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"https:\/\/courses.lumenlearning.com\/waymakercollegealgebracorequisite\/\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Precalculus\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"https:\/\/courses.lumenlearning.com\/precalculus\/\",\"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-1974","chapter","type-chapter","status-publish","hentry"],"part":1968,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1974","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\/349141"}],"version-history":[{"count":1,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1974\/revisions"}],"predecessor-version":[{"id":2460,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1974\/revisions\/2460"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/1968"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1974\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=1974"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=1974"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=1974"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=1974"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}