Evaluate the integral. (Use C for the constant of integration.) 8 tan\u00b3(x) sec(x) dx 8 sec\u00b3(x) 3-sec(x) +C<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n To evaluate the integral \u222b8tan\u20613(x)sec\u2061(x)\u2009dx\\int 8 \\tan^3(x) \\sec(x) \\, dx, let’s break it down step by step.<\/p>\n\n\n\n We know the relationships between tan\u2061(x)\\tan(x) and sec\u2061(x)\\sec(x): sec\u20612(x)=1+tan\u20612(x)\\sec^2(x) = 1 + \\tan^2(x) ddx[tan\u2061(x)]=sec\u20612(x)\\frac{d}{dx}[\\tan(x)] = \\sec^2(x)<\/p>\n\n\n\n This suggests a substitution involving tan\u2061(x)\\tan(x).<\/p>\n\n\n\n Let: u=tan\u2061(x)u = \\tan(x)<\/p>\n\n\n\n Then: du=sec\u20612(x)\u2009dxdu = \\sec^2(x) \\, dx<\/p>\n\n\n\n Also, since sec\u2061(x)=1+u2\\sec(x) = \\sqrt{1 + u^2}, we can rewrite the given integral in terms of uu.<\/p>\n\n\n\n The given integral becomes: 8\u222btan\u20613(x)sec\u2061(x)\u2009dx=8\u222bu31+u2\u2009du8 \\int \\tan^3(x) \\sec(x) \\, dx = 8 \\int u^3 \\sqrt{1 + u^2} \\, du<\/p>\n\n\n\n Expand the integral: \u222bu31+u2\u2009du\\int u^3 \\sqrt{1 + u^2} \\, du<\/p>\n\n\n\n We use substitution to handle the square root. Let: v=1+u2so thatdv=2u\u2009duv = 1 + u^2 \\quad \\text{so that} \\quad dv = 2u \\, du<\/p>\n\n\n\n Substitute u2=v\u22121u^2 = v – 1 and u\u2009du=12dvu \\, du = \\frac{1}{2} dv, and the integral becomes: 8\u222bu31+u2\u2009du=8\u222b(v\u22121)v\u22c512dv8 \\int u^3 \\sqrt{1 + u^2} \\, du = 8 \\int (v – 1) \\sqrt{v} \\cdot \\frac{1}{2} dv<\/p>\n\n\n\n Simplify: 4\u222b(v3\/2\u2212v1\/2)\u2009dv4 \\int (v^{3\/2} – v^{1\/2}) \\, dv<\/p>\n\n\n\n Use the power rule for each term: \u222bv3\/2\u2009dv=25v5\/2,\u222bv1\/2\u2009dv=23v3\/2\\int v^{3\/2} \\, dv = \\frac{2}{5} v^{5\/2}, \\quad \\int v^{1\/2} \\, dv = \\frac{2}{3} v^{3\/2}<\/p>\n\n\n\n So: 4\u222b(v3\/2\u2212v1\/2)\u2009dv=4(25v5\/2\u221223v3\/2)4 \\int (v^{3\/2} – v^{1\/2}) \\, dv = 4 \\left( \\frac{2}{5} v^{5\/2} – \\frac{2}{3} v^{3\/2} \\right)<\/p>\n\n\n\n Replace v=1+u2v = 1 + u^2 and u=tan\u2061(x)u = \\tan(x): =85(1+tan\u20612(x))5\/2\u221283(1+tan\u20612(x))3\/2+C= \\frac{8}{5} (1 + \\tan^2(x))^{5\/2} – \\frac{8}{3} (1 + \\tan^2(x))^{3\/2} + C<\/p>\n\n\n\n The evaluated integral is: 85sec\u20615(x)\u221283sec\u20613(x)+C\\frac{8}{5} \\sec^5(x) – \\frac{8}{3} \\sec^3(x) + C<\/p>\n\n\n\n This matches the expression in terms of sec\u2061(x)\\sec(x).<\/p>\n","protected":false},"excerpt":{"rendered":" Evaluate the integral. (Use C for the constant of integration.) 8 tan\u00b3(x) sec(x) dx 8 sec\u00b3(x) 3-sec(x) +C The Correct Answer and Explanation is : To evaluate the integral \u222b8tan\u20613(x)sec\u2061(x)\u2009dx\\int 8 \\tan^3(x) \\sec(x) \\, dx, let’s break it down step by step. Step 1: Simplify the expression We know the relationships between tan\u2061(x)\\tan(x) and sec\u2061(x)\\sec(x): […]<\/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-185413","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/185413","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=185413"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/185413\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=185413"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=185413"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=185413"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Step 1: Simplify the expression<\/h3>\n\n\n\n
Step 2: Substitution<\/h3>\n\n\n\n
Step 3: Rewrite the integral<\/h3>\n\n\n\n
Step 4: Simplify further<\/h3>\n\n\n\n
Step 5: Integrate<\/h3>\n\n\n\n
Step 6: Substitute back<\/h3>\n\n\n\n
Final Answer:<\/h3>\n\n\n\n