{"id":8595,"date":"2023-12-31T19:36:38","date_gmt":"2023-12-31T14:06:38","guid":{"rendered":"https:\/\/physicscatalyst.com\/article\/?p=8595"},"modified":"2024-01-17T23:18:00","modified_gmt":"2024-01-17T17:48:00","slug":"integration-of-trigonometric-functions","status":"publish","type":"post","link":"https:\/\/physicscatalyst.com\/article\/integration-of-trigonometric-functions\/","title":{"rendered":"Integration of trigonometric functions"},"content":{"rendered":"\n<p>Integrating trigonometric functions often involves using various integration techniques, including basic integration formulas, substitution, integration by parts, and trigonometric identities. Here&#8217;s a brief overview of some common integrals of trigonometric functions:<\/p>\n\n\n\n<h2 class=\"wp-block-heading\">Basic Trigonometric Integrals<\/h2>\n\n\n\n<ul class=\"wp-block-list\">\n<li>$\\int (\\cos x) = \\sin x + C$<\/li>\n\n\n\n<li>$\\int (\\sin x) = &#8211; \\cos x + C$<\/li>\n\n\n\n<li>$\\int ( \\sec^2 x) = \\tan x + C$<\/li>\n\n\n\n<li>$\\int (\\csc^2 x) = -\\cot x + C$<\/li>\n\n\n\n<li>$\\int ( \\sec (x) \\tan (x) )=\\sec x + C$<\/li>\n\n\n\n<li>$\\int (\\csc (x) \\cot (x)) = -\\csc x + C$<\/li>\n<\/ul>\n\n\n\n<h2 class=\"wp-block-heading\">Integrals of Tangent and Cotangent<\/h2>\n\n\n\n<p>This can be solved using Integration by substitution<\/p>\n\n\n\n<p>(I)<\/p>\n\n\n\n<p>\\[<br>\\int \\cot(x) \\, dx = \\int  \\frac {\\cos(x)}{\\sin(x)}  \\, dx<br>\\]<\/p>\n\n\n\n<p>Now <br>$\\sin(x) =t$<br>then<br>$\\cos (x) dx = dt$<br>Therefore,<\/p>\n\n\n\n<p>\\[<br>\\int \\cot(x) \\, dx = \\int  \\frac {1}{t}  \\, dt = \\ln |sin(x)| + C<br>\\]<\/p>\n\n\n\n<p>(II)<\/p>\n\n\n\n<p>\\[<br>\\int \\tan(x) \\, dx = \\int  \\frac {\\sin(x)}{\\cos(x)}  \\, dx<br>\\]<\/p>\n\n\n\n<p>Now <br>$\\cos(x) =t$<br>then<br>$-\\sin (x) dx = dt$<br>Therefore,<\/p>\n\n\n\n<p>\\[<br>\\int \\cot(x) \\, dx = -\\int  \\frac {1}{t}  \\, dt = -\\ln |cos(x)| + C=ln|\\sec (x)| + C<br>\\]<\/p>\n\n\n\n<h2 class=\"wp-block-heading\">Integrals of Secant and Cosecant<\/h2>\n\n\n\n<p>(I)<\/p>\n\n\n\n<p>Integral of cosecant can be little tricky and can be solved by multiplying and dividing by $(\\csc(x) &#8211; \\cot(x))$. Here&#8217;s how it&#8217;s done:<\/p>\n\n\n\n<p>\\[<br> \\int \\csc(x) \\, dx = \\int \\csc(x) \\cdot \\frac{\\csc(x) &#8211; \\cot(x)}{\\csc(x) -\\cot(x)} \\, dx \\\\ <br>= \\int \\frac{\\csc^2(x) &#8211; \\csc(x)\\cot(x)}{\\csc(x) &#8211; \\cot(x)} \\, dx. <br>\\]<\/p>\n\n\n\n<p>Now,  taking u as<br>$ u= \\csc(x) &#8211; \\cot(x) $<br>$ du =(-\\csc(x)\\cot(x) + \\csc^2(x)) dx $<\/p>\n\n\n\n<p>Therefore integration becomes<\/p>\n\n\n\n<p>\\[<br>\\int \\csc(x) \\, dx =\\int \\frac {1}{u} \\, du,<br>\\]<\/p>\n\n\n\n<p>hence<\/p>\n\n\n\n<p>\\[<br>\\int \\csc(x) \\, dx = \\ln | \\csc(x) &#8211; \\cot(x) | + C<br>\\]<\/p>\n\n\n\n<p>(II) <\/p>\n\n\n\n<p>Integral of secant can be little tricky and can be solved by multiplying and dividing by $(\\sec(x) + \\tan(x))$. Here&#8217;s how it&#8217;s done:<\/p>\n\n\n\n<p>\\[<br> \\int \\sec(x) \\, dx = \\int \\sec(x) \\cdot \\frac{\\sec(x) + \\tan(x)}{\\sec(x) + \\tan(x)} \\, dx \\\\ <br>= \\int \\frac{\\sec^2(x) + \\sec(x)\\tan(x)}{\\sec(x) + \\tan(x)} \\, dx. <br>\\]<\/p>\n\n\n\n<p>Now,  taking u as<br>$ u= \\sec(x) + \\tan(x) $<br>$ du =(\\sec(x)\\tan(x) + \\sec^2(x)) dx $<\/p>\n\n\n\n<p>Therefore integration becomes<\/p>\n\n\n\n<p>\\[<br>\\int \\sec(x) \\, dx =\\int \\frac {1}{u} \\, du,<br>\\]<\/p>\n\n\n\n<p>hence<\/p>\n\n\n\n<p>\\[<br>\\int \\sec(x) \\, dx = \\ln | \\sec(x) + \\tan(x) | + C<br>\\]<\/p>\n\n\n\n<h2 class=\"wp-block-heading\">Inverse Trigonometric Functions<\/h2>\n\n\n\n<p>Integrals that result in inverse trigonometric functions, like $\\int \\frac{dx}{\\sqrt{1 &#8211; x^2}} = \\sin^{-1}(x) + C$ or $\\int \\frac{dx}{1 + x^2} = \\tan^{-1}(x) + C$.<\/p>\n\n\n\n<p>$\\int ( \\frac {1}{\\sqrt {1-x^2} } ) = \\sin^{-1}x + C$<\/p>\n\n\n\n<p>$\\int (\\frac {1}{\\sqrt {1-x^2}}) = &#8211; \\cos ^{-1}x&nbsp; +C$<\/p>\n\n\n\n<p>$\\int ( \\frac {1}{1 + x^2}) =\\tan ^{-1}x + C$<\/p>\n\n\n\n<p>$\\int ( \\frac {1}{1 + x^2}) = -\\cot ^{-1}x + C$<\/p>\n\n\n\n<p>$\\int (\\frac {1}{|x|\\sqrt {x^-1}}) = -sec^{-1} x + C $<\/p>\n\n\n\n<p>$\\int (\\frac {1}{|x|\\sqrt {x^-1}}) = -cosec^{-1} x + C $<\/p>\n\n\n\n<h2 class=\"wp-block-heading\">Integrals Involving Substitution:<\/h2>\n\n\n\n<p>For integrals like $\\int \\sin(ax+b) \\, dx$ or $\\int \\cos(ax + b) \\, dx$, a substitution such as (u = ax+b)  can be used.<\/p>\n\n\n\n<p>$\\int \\cos (ax+b) = \\frac {1}{a}&nbsp; \\sin (ax+b) + C$<\/p>\n\n\n\n<p>$\\int \\sin (ax+b) = &#8211; \\frac {1}{a} \\cos (ax+b) + C$<\/p>\n\n\n\n<p>$\\int \\sec^2 (ax+b) = \\frac {1}{a}&nbsp; \\tan (ax +b) + C$<\/p>\n\n\n\n<p>$\\int \\csc^2 (ax+b) = &#8211; \\frac {1}{a}&nbsp; \\cot (ax+b)+ C$<\/p>\n\n\n\n<p>$ \\int \\tan (ax+b) =- \\frac {1}{a}&nbsp; ln |\\cos (ax+b)| + C$<\/p>\n\n\n\n<p>$ \\int \\cot (ax+b) = \\frac {1}{a}&nbsp; ln |\\sin (ax+b)| + C$<\/p>\n\n\n\n<p>$ \\int \\sec (ax+b) =\\frac {1}{a} ln |\\sec (ax+b) + \\tan (ax+b)| + C$<\/p>\n\n\n\n<p>$ \\int \\csc (ax+b) = \\frac {1}{a} ln |\\csc (ax+b) &#8211; \\cot (ax+b)| + C$<\/p>\n\n\n\n<h2 class=\"wp-block-heading\">Integrals Involving Trigonometric Identities<\/h2>\n\n\n\n<p>Sometimes, using trigonometric identities to simplify the integrand is helpful. For example, $\\sin^2(x)$ can be expressed as$\\frac{1 &#8211; \\cos(2x)}{2}$ using the double-angle identity<\/p>\n\n\n\n<p>(I)<\/p>\n\n\n\n<p>$\\int \\sin^2 x \\; dx$ or $\\int \\cos^2 x \\; dx$<br>We know that<br>$\\cos2x = \\cos^{2}x-\\sin^{2}x = 2\\cos^{2} x-1=1-2\\sin^{2}x$<br>So<br>$\\cos^2 x= \\frac {1+ \\cos 2x}{2}$<br>$\\sin^2 x= \\frac {1- \\cos 2x}{2}$<\/p>\n\n\n\n<p>(II)<\/p>\n\n\n\n<p>$\\int \\sin^3 x \\; dx$ or $\\int \\cos^3 x \\; dx$<br>We know that<br>$\\sin(3x)=3\\sin(x)-4\\sin^{3}x$<br>$\\sin^{3}x= \\frac { 3 \\sin x &#8211; \\sin 3x}{4}$<br>Also<br>$\\cos(3x)=4\\cos^{3}x-3\\cos(x)$<br>$\\cos^{3}x= \\frac {\\cos 3x + 3 \\cos x}{4}$<\/p>\n\n\n\n<p>(III)<\/p>\n\n\n\n<p>$\\int \\tan^2 x \\; dx$ or $\\int \\cot^2 x \\; dx$<br>We know that<br>$\\sec^2x =1 + \\tan^2x$<br>or $\\tan^2x =\\sec^2 x -1$<br>$\\int \\tan^2 x dx = \\int (\\sec^2 x -1) dx$<br>Now $\\int ( \\sec^2 x) \\; dx = \\tan x + C$<br>Therefore<br>$\\int \\tan^2 x \\; dx = \\int (\\sec^2 x -1) dx= \\tan x -x + C$<br>We know that<br>$cosec^2 x =1 + \\cot^2 x$<br>or $ \\cot^2 x = cosec^2 x -1$<br>$\\int \\cot^2 x \\; dx = \\int (cosec^2 x -1) \\; dx$<br>Now $\\int ( cosec^2 x) \\; dx = -\\cot x + C$<br>Therefore<br>$\\int \\cot^2 x \\; dx = \\int (cosec^2 x -1) \\; dx= -\\cot x -x + C$<\/p>\n\n\n\n<h2 class=\"wp-block-heading\">Integration by Parts<\/h2>\n\n\n\n<p>This technique is useful for integrals like $\\int x \\sin(x) \\, dx$ or $\\int x \\cos(x) \\, dx$<\/p>\n\n\n\n<p>$\\int x \\sin(x) \\; dx = \\sin x -x \\cos(x) + C$<\/p>\n\n\n\n<p>$\\int x \\cos(x) \\; dx = \\cos x +x \\sin(x) + C$<\/p>\n\n\n\n<p>When integrating trigonometric functions, it&#8217;s often helpful to be familiar with various trigonometric identities and properties, as they can simplify the integration process significantly. Additionally, the choice of technique depends on the specific form of the integrand.<\/p>\n\n\n\n<div class=\"wp-block-group has-ast-global-color-4-background-color has-background is-layout-constrained wp-container-core-group-is-layout-ca99af60 wp-block-group-is-layout-constrained\" style=\"border-radius:17px;margin-top:var(--wp--preset--spacing--20);margin-bottom:var(--wp--preset--spacing--20);padding-top:var(--wp--preset--spacing--20);padding-right:var(--wp--preset--spacing--50);padding-bottom:var(--wp--preset--spacing--20);padding-left:var(--wp--preset--spacing--50)\"><div class=\"wp-block-group__inner-container\">\n<p class=\"has-medium-font-size\">Other Integration Related Articles<\/p>\n\n\n\n<figure class=\"wp-block-table is-style-regular\" style=\"padding-right:0;padding-left:0;font-size:16px\"><table style=\"border-style:none;border-width:0px\"><tbody><tr><td><a href=\"https:\/\/physicscatalyst.com\/article\/integration-of-tan-inverse-x\/\">Integration of tan inverse x<\/a><\/td><td><a href=\"https:\/\/physicscatalyst.com\/article\/integration-of-cos-inverse-x\/\">Integration of cos inverse x<\/a><\/td><\/tr><tr><td><a href=\"https:\/\/physicscatalyst.com\/article\/integration-of-cos-square-x\/\">Integration of cos square x<\/a><\/td><td><a href=\"https:\/\/physicscatalyst.com\/article\/integration-of-fractional-part-of-x\/\">Integration of fractional part of x<\/a><\/td><\/tr><tr><td><a href=\"https:\/\/physicscatalyst.com\/article\/integration-of-uv-formula\/\">integration of uv formula<\/a><\/td><td><a href=\"https:\/\/physicscatalyst.com\/article\/integration-of-log-x\/\">integration of log x<\/a><\/td><\/tr><tr><td><a href=\"https:\/\/physicscatalyst.com\/article\/integration-of-sin-inverse-x\/\">integration of sin inverse x<\/a><\/td><td><a href=\"https:\/\/physicscatalyst.com\/article\/integration-of-constant\/\">Integration of constant term<\/a><\/td><\/tr><tr><td><a href=\"https:\/\/physicscatalyst.com\/article\/integration-of-hyperbolic-functions\/\">integration of hyperbolic functions<\/a><\/td><td><a data-type=\"post\" data-id=\"5701\" href=\"https:\/\/physicscatalyst.com\/article\/integration-formulas-list\/\">List of Integration Formulas <\/a><\/td><\/tr><tr><td><a href=\"https:\/\/physicscatalyst.com\/article\/integration-of-log-sinx\/\">integration of log sinx<\/a><\/td><td><a href=\"https:\/\/physicscatalyst.com\/article\/integration-of-sin-x-cos-x-dx\/\">integration of sin x cos x dx<\/a><\/td><\/tr><\/tbody><\/table><\/figure>\n<\/div><\/div>\n","protected":false},"excerpt":{"rendered":"<p>Integrating trigonometric functions often involves using various integration techniques, including basic integration formulas, substitution, integration by parts, and trigonometric identities. Here&#8217;s a brief overview of some common integrals of trigonometric functions: Basic Trigonometric Integrals Integrals of Tangent and Cotangent This can be solved using Integration by substitution (I) \\[\\int \\cot(x) \\, dx = \\int \\frac [&hellip;]<\/p>\n","protected":false},"author":8,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_uag_custom_page_level_css":"","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":"","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":"set","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":[498],"tags":[506],"class_list":["post-8595","post","type-post","status-publish","format-standard","hentry","category-maths","tag-integration"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v27.4 - https:\/\/yoast.com\/product\/yoast-seo-wordpress\/ -->\n<title>Integration of trigonometric functions - physicscatalyst&#039;s Blog<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/physicscatalyst.com\/article\/integration-of-trigonometric-functions\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Integration of trigonometric functions - physicscatalyst&#039;s Blog\" \/>\n<meta property=\"og:description\" content=\"Integrating trigonometric functions often involves using various integration techniques, including basic integration formulas, substitution, integration by parts, and trigonometric identities. 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Here&#8217;s a brief overview of some common integrals of trigonometric functions: Basic Trigonometric Integrals Integrals of Tangent and Cotangent This can be solved using Integration by substitution (I) [int cot(x) , dx = int frac [&hellip;]","og_url":"https:\/\/physicscatalyst.com\/article\/integration-of-trigonometric-functions\/","og_site_name":"physicscatalyst&#039;s Blog","article_publisher":"https:\/\/www.facebook.com\/PhysicsCatalyst","article_author":"https:\/\/www.facebook.com\/PhysicsCatalyst","article_published_time":"2023-12-31T14:06:38+00:00","article_modified_time":"2024-01-17T17:48:00+00:00","author":"physicscatalyst","twitter_misc":{"Written by":"physicscatalyst","Est. reading time":"4 minutes"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"Article","@id":"https:\/\/physicscatalyst.com\/article\/integration-of-trigonometric-functions\/#article","isPartOf":{"@id":"https:\/\/physicscatalyst.com\/article\/integration-of-trigonometric-functions\/"},"author":{"name":"physicscatalyst","@id":"https:\/\/physicscatalyst.com\/article\/#\/schema\/person\/9b302efdc9b32e459cb1e61ab7506d3f"},"headline":"Integration of trigonometric functions","datePublished":"2023-12-31T14:06:38+00:00","dateModified":"2024-01-17T17:48:00+00:00","mainEntityOfPage":{"@id":"https:\/\/physicscatalyst.com\/article\/integration-of-trigonometric-functions\/"},"wordCount":869,"commentCount":0,"publisher":{"@id":"https:\/\/physicscatalyst.com\/article\/#organization"},"keywords":["Integration"],"articleSection":["Maths"],"inLanguage":"en-US","potentialAction":[{"@type":"CommentAction","name":"Comment","target":["https:\/\/physicscatalyst.com\/article\/integration-of-trigonometric-functions\/#respond"]}]},{"@type":"WebPage","@id":"https:\/\/physicscatalyst.com\/article\/integration-of-trigonometric-functions\/","url":"https:\/\/physicscatalyst.com\/article\/integration-of-trigonometric-functions\/","name":"Integration of trigonometric functions - 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Here&#8217;s a brief overview of some common integrals of trigonometric functions: Basic Trigonometric Integrals Integrals of Tangent and Cotangent This can be solved using Integration by substitution (I) \\[\\int \\cot(x) \\, dx = \\int \\frac&hellip;","_links":{"self":[{"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/posts\/8595","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/users\/8"}],"replies":[{"embeddable":true,"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/comments?post=8595"}],"version-history":[{"count":3,"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/posts\/8595\/revisions"}],"predecessor-version":[{"id":8802,"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/posts\/8595\/revisions\/8802"}],"wp:attachment":[{"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/media?parent=8595"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/categories?post=8595"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/tags?post=8595"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}