{"id":231624,"date":"2025-06-11T08:57:21","date_gmt":"2025-06-11T08:57:21","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=231624"},"modified":"2025-06-11T08:57:23","modified_gmt":"2025-06-11T08:57:23","slug":"evaluate-the-indefinite-integral-by-using-the-given-substitution-to-reduce-the-integral-to-standard-form","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/11\/evaluate-the-indefinite-integral-by-using-the-given-substitution-to-reduce-the-integral-to-standard-form\/","title":{"rendered":"Evaluate the indefinite integral by using the given substitution to reduce the integral to standard form."},"content":{"rendered":"\n

Evaluate the indefinite integral by using the given substitution to reduce the integral to standard form.\\
, a.
, b.
\\ a. Using
,
\\ b. Using
,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

We are given the integral:\u222b6csc\u20612(3x)cot\u2061(3x)\u2009dx\\int 6 \\csc^2(3x) \\cot(3x)\\, dx\u222b6csc2(3x)cot(3x)dx<\/p>\n\n\n\n

We are to evaluate it using two different substitutions:<\/p>\n\n\n\n

\n\n\n\n

a. Using u=cot\u2061(3x)u = \\cot(3x)u=cot(3x):<\/strong><\/h3>\n\n\n\n

Step 1: Differentiate u=cot\u2061(3x)u = \\cot(3x)u=cot(3x)dudx=\u22123csc\u20612(3x)\u21d2du=\u22123csc\u20612(3x)\u2009dx\\frac{du}{dx} = -3 \\csc^2(3x) \\quad \\Rightarrow \\quad du = -3 \\csc^2(3x) \\, dxdxdu\u200b=\u22123csc2(3x)\u21d2du=\u22123csc2(3x)dx<\/p>\n\n\n\n

Solve for csc\u20612(3x)\u2009dx\\csc^2(3x)\\, dxcsc2(3x)dx:csc\u20612(3x)\u2009dx=\u221213\u2009du\\csc^2(3x)\\, dx = -\\frac{1}{3} \\, ducsc2(3x)dx=\u221231\u200bdu<\/p>\n\n\n\n

Now substitute into the integral:\u222b6csc\u20612(3x)cot\u2061(3x)\u2009dx=\u222b6cot\u2061(3x)\u22c5csc\u20612(3x)\u2009dx\\int 6 \\csc^2(3x) \\cot(3x) \\, dx = \\int 6 \\cot(3x) \\cdot \\csc^2(3x) \\, dx\u222b6csc2(3x)cot(3x)dx=\u222b6cot(3x)\u22c5csc2(3x)dx<\/p>\n\n\n\n

Substitute cot\u2061(3x)=u\\cot(3x) = ucot(3x)=u, and csc\u20612(3x)\u2009dx=\u221213du\\csc^2(3x)\\, dx = -\\frac{1}{3} ducsc2(3x)dx=\u221231\u200bdu:=\u222b6u\u22c5(\u221213)\u2009du=\u22122\u222bu\u2009du=\u22122\u22c5u22+C=\u2212u2+C= \\int 6u \\cdot \\left(-\\frac{1}{3}\\right) \\, du = -2 \\int u \\, du = -2 \\cdot \\frac{u^2}{2} + C = -u^2 + C=\u222b6u\u22c5(\u221231\u200b)du=\u22122\u222budu=\u22122\u22c52u2\u200b+C=\u2212u2+C<\/p>\n\n\n\n

Substitute back u=cot\u2061(3x)u = \\cot(3x)u=cot(3x):\u2212cot\u20612(3x)+C\\boxed{-\\cot^2(3x) + C}\u2212cot2(3x)+C\u200b<\/p>\n\n\n\n

\n\n\n\n

b. Using u=csc\u2061(3x)u = \\csc(3x)u=csc(3x):<\/strong><\/h3>\n\n\n\n

Step 1: Differentiate u=csc\u2061(3x)u = \\csc(3x)u=csc(3x)dudx=\u22123csc\u2061(3x)cot\u2061(3x)\u21d2du=\u22123csc\u2061(3x)cot\u2061(3x)\u2009dx\\frac{du}{dx} = -3 \\csc(3x) \\cot(3x) \\quad \\Rightarrow \\quad du = -3 \\csc(3x) \\cot(3x) \\, dxdxdu\u200b=\u22123csc(3x)cot(3x)\u21d2du=\u22123csc(3x)cot(3x)dx<\/p>\n\n\n\n

Solve for csc\u2061(3x)cot\u2061(3x)dx\\csc(3x) \\cot(3x) dxcsc(3x)cot(3x)dx:csc\u2061(3x)cot\u2061(3x)\u2009dx=\u221213\u2009du\\csc(3x) \\cot(3x)\\, dx = -\\frac{1}{3} \\, ducsc(3x)cot(3x)dx=\u221231\u200bdu<\/p>\n\n\n\n

Now write the integral in terms of uuu:\u222b6csc\u20612(3x)cot\u2061(3x)\u2009dx=\u222b6csc\u2061(3x)\u22c5csc\u2061(3x)cot\u2061(3x)\u2009dx\\int 6 \\csc^2(3x) \\cot(3x) \\, dx = \\int 6 \\csc(3x) \\cdot \\csc(3x) \\cot(3x) \\, dx\u222b6csc2(3x)cot(3x)dx=\u222b6csc(3x)\u22c5csc(3x)cot(3x)dx<\/p>\n\n\n\n

Group: csc\u2061(3x)cot\u2061(3x)\u2009dx=\u221213du\\csc(3x) \\cot(3x)\\, dx = -\\frac{1}{3} ducsc(3x)cot(3x)dx=\u221231\u200bdu, so:=\u222b6csc\u2061(3x)\u22c5(\u221213)\u2009du=\u22122\u222bu\u2009du=\u2212u2+C= \\int 6 \\csc(3x) \\cdot (-\\frac{1}{3}) \\, du = -2 \\int u \\, du = -u^2 + C=\u222b6csc(3x)\u22c5(\u221231\u200b)du=\u22122\u222budu=\u2212u2+C<\/p>\n\n\n\n

Substitute back u=csc\u2061(3x)u = \\csc(3x)u=csc(3x):\u2212csc\u20612(3x)+C\\boxed{-\\csc^2(3x) + C}\u2212csc2(3x)+C\u200b<\/p>\n\n\n\n

\n\n\n\n

\ud83d\udcd8 Explanation<\/strong><\/h3>\n\n\n\n

To evaluate the integral \u222b6csc\u20612(3x)cot\u2061(3x)\u2009dx\\int 6 \\csc^2(3x) \\cot(3x)\\, dx\u222b6csc2(3x)cot(3x)dx, we use u-substitution<\/strong>, a technique that simplifies a complex expression into a standard integral form.<\/p>\n\n\n\n

In part (a)<\/strong>, we let u=cot\u2061(3x)u = \\cot(3x)u=cot(3x), a function whose derivative involves csc\u20612(3x)\\csc^2(3x)csc2(3x), which is present in the integrand. By differentiating uuu, we find du=\u22123csc\u20612(3x)dxdu = -3 \\csc^2(3x) dxdu=\u22123csc2(3x)dx. This tells us how to express part of the integrand in terms of dududu. When we rearrange, we get csc\u20612(3x)dx=\u221213du\\csc^2(3x) dx = -\\frac{1}{3} ducsc2(3x)dx=\u221231\u200bdu, and substitute into the integral. The original expression becomes \u222b6u(\u221213)du=\u22122\u222bu\u2009du\\int 6u(-\\frac{1}{3}) du = -2 \\int u\\, du\u222b6u(\u221231\u200b)du=\u22122\u222budu, a basic polynomial integral. Solving yields \u2212u2+C-u^2 + C\u2212u2+C, and substituting back gives \u2212cot\u20612(3x)+C-\\cot^2(3x) + C\u2212cot2(3x)+C.<\/p>\n\n\n\n

In part (b)<\/strong>, we instead use u=csc\u2061(3x)u = \\csc(3x)u=csc(3x), another trigonometric identity involved in the integrand. Differentiating gives du=\u22123csc\u2061(3x)cot\u2061(3x)dxdu = -3 \\csc(3x) \\cot(3x) dxdu=\u22123csc(3x)cot(3x)dx, and solving gives csc\u2061(3x)cot\u2061(3x)dx=\u221213du\\csc(3x)\\cot(3x) dx = -\\frac{1}{3} ducsc(3x)cot(3x)dx=\u221231\u200bdu. The original integral is rewritten as \u222b6csc\u2061(3x)\u22c5csc\u2061(3x)cot\u2061(3x)dx=\u222b6csc\u2061(3x)\u22c5(\u221213)du\\int 6 \\csc(3x) \\cdot \\csc(3x) \\cot(3x) dx = \\int 6 \\csc(3x) \\cdot (-\\frac{1}{3}) du\u222b6csc(3x)\u22c5csc(3x)cot(3x)dx=\u222b6csc(3x)\u22c5(\u221231\u200b)du. This again reduces to a simple polynomial form: \u22122\u222bu\u2009du=\u2212u2+C-2 \\int u\\, du = -u^2 + C\u22122\u222budu=\u2212u2+C. Replacing uuu yields the final result: \u2212csc\u20612(3x)+C-\\csc^2(3x) + C\u2212csc2(3x)+C.<\/p>\n\n\n\n

Each method is valid and demonstrates the power of substitution for integrating complex trigonometric expressions.<\/p>\n\n\n\n

\"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

Evaluate the indefinite integral by using the given substitution to reduce the integral to standard form.\\, a., b.\\ a. Using,\\ b. Using, The Correct Answer and Explanation is: We are given the integral:\u222b6csc\u20612(3x)cot\u2061(3x)\u2009dx\\int 6 \\csc^2(3x) \\cot(3x)\\, dx\u222b6csc2(3x)cot(3x)dx We are to evaluate it using two different substitutions: a. Using u=cot\u2061(3x)u = \\cot(3x)u=cot(3x): Step 1: Differentiate u=cot\u2061(3x)u […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-231624","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/231624","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=231624"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/231624\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=231624"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=231624"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=231624"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}