{"id":8767,"date":"2024-01-15T23:52:50","date_gmt":"2024-01-15T18:22:50","guid":{"rendered":"https:\/\/physicscatalyst.com\/article\/?p=8767"},"modified":"2024-01-29T00:16:45","modified_gmt":"2024-01-28T18:46:45","slug":"integration-of-tan-inverse-x","status":"publish","type":"post","link":"https:\/\/physicscatalyst.com\/article\/integration-of-tan-inverse-x\/","title":{"rendered":"Integration of tan inverse x"},"content":{"rendered":"\n<p>Integration of tan inverse x can be calculated using integration by parts  .Here is the formula for it<\/p>\n\n\n\n<p>\\[ \\int \\tan^{-1}x \\, dx = x \\tan^{-1}x &#8211; \\frac {1}{2} \\ln (1 + x^2) + C  \\]<\/p>\n\n\n\n<p>where (C) is the constant of integration.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\">Proof of integration of tan inverse x<\/h2>\n\n\n\n<p>To find the integral  we use integration by parts. Integration by parts is based on the product rule for differentiation and is given by:<\/p>\n\n\n\n<p>$\\int f(x) g(x) dx = f(x) (\\int g(x) dx )- \\int \\left \\{ \\frac {df(x)}{dx} \\int g(x) dx \\right \\} dx $<br>In our case, we can let $ f(x) = \\tan^{-1}x $ and $ g(x) = 1 $. Then<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>$ \\frac {df(x)}{dx} = \\frac{1}{1+x^2} \\, dx $ <\/li>\n\n\n\n<li>$\\int g(x) dx = \\int dx = x $<\/li>\n<\/ul>\n\n\n\n<p>Now, substitute these into the integration by parts formula:<\/p>\n\n\n\n<p>$$<br>\\int \\tan^{-1}x \\, dx  = x \\tan^{-1}x  &#8211; \\int x \\cdot \\frac{1}{1+x^2}  \\, dx<br>$$<\/p>\n\n\n\n<p>$ =x \\tan^{-1}x  &#8211; \\int  \\frac{x}{1+x^2}  \\, dx \\\\<br>  =x \\tan^{-1}x  &#8211; \\frac {1}{2} \\int  \\frac{2x}{1+x^2}  \\, dx $<\/p>\n\n\n\n<p>Now lets calculate the second integral separately  $\\int  \\frac{2x}{1+x^2}  \\, dx $<\/p>\n\n\n\n<p>Let $t= 1+x^2$<br>then $dt=2x dx$<br>Therefore<\/p>\n\n\n\n<p>$\\int  \\frac{2x}{1+x^2}  \\, dx  = \\int  \\frac{1}{t}  \\, dx  \\\\<br>      = \\ln |t|  = \\ln |1+x^2|$<\/p>\n\n\n\n<p>Substituting this value in main integral , we get<\/p>\n\n\n\n<p>$$<br>\\int \\tan^{-1}x \\, dx  = x \\tan^{-1}x   &#8211; \\frac {1}{2} \\ln (1 + x^2) + C  $$<\/p>\n\n\n\n<h2 class=\"wp-block-heading\">Definite Integral of tan inverse x<\/h2>\n\n\n\n<p>To find the definite integral of $\\tan^{-1}x$  over a specific interval, we use the same approach as with the indefinite integral, but we&#8217;ll apply the limits of integration at the end.<\/p>\n\n\n\n<p>The definite integral of $\\tan^{-1}x$ from $a$ to $b$ is given by:<\/p>\n\n\n\n<p>$$\\int_{a}^{b} \\tan^{-1}x \\, dx = b \\tan^{-1}b &#8211; a \\cos^{-1}a + \\frac {1}{2}\\ln (1 + a^2) &#8211; \\frac {1}{2} \\ln (1 + b^2) $$<\/p>\n\n\n\n<p>This expression represents the accumulated area under the curve of $\\tan^{-1}x$ from $x = a$ to $x = b$.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\">Solved Examples on Integration of tan inverse x<\/h2>\n\n\n\n<p><strong>Question 1<\/strong><\/p>\n\n\n\n<p>$$\\int_{0}^{1} \\tan^{-1}x  \\, dx$$<\/p>\n\n\n\n<p><strong>Solution<\/strong><\/p>\n\n\n\n<p>$\\int_{0}^{1} \\tan^{-1}x  \\, dx = [x \\tan^{-1}x &#8211; \\frac {1}{2} \\ln (1 + x^2)] _0 ^1 \\\\<br>      = 1. tan^{-1} 1 &#8211; \\frac {\\ln 2}{2}  &#8211; 0 + \\frac {\\ln 1}{2} = \\frac {\\pi}{4} &#8211; \\ln 2\/2$<\/p>\n\n\n\n<p><strong>Question  2<\/strong><\/p>\n\n\n\n<p>$$ \\int x tan^{-1} x  \\, dx$$<\/p>\n\n\n\n<p><strong>Solution<\/strong><\/p>\n\n\n\n<p>Applying integration by parts<br>$\\int u \\, dv = uv &#8211; \\int v \\, du$.<br>Let $u = \\tan^{-1}x$. Then, $du = \\frac{dx}{1 + x^2}$.<br>Let $dv = x \\, dx$. Then, $v = \\frac{x^2}{2}$.<br>Then<br>$ \\int x \\tan^{-1}x \\, dx = \\frac{x^2}{2} \\tan^{-1}x &#8211; \\int \\frac{x^2}{2} \\cdot \\frac{dx}{1 + x^2} \\<br>=\\frac{x^2}{2} \\tan^{-1}x &#8211; \\frac {1}{2} \\int \\frac{x^2}{1 + x^2} \\; dx$<br>Now<br>$\\int \\frac{x^2}{1 + x^2} \\; dx = \\int \\frac{1+x^2 &#8211; 1}{1 + x^2} \\; dx \\\\<br>= \\int 1 dx &#8211; \\int \\frac {1}{1+x^2} dx \\\\<br>= x &#8211; tan^{-}x$<\/p>\n\n\n\n<p>Applying these steps to the integral of $x \\tan^{-1}x$, we obtained:<\/p>\n\n\n\n<p>$$ \\frac{x^2 \\tan^{-1}x}{2} &#8211; \\frac{x }{2} + \\frac{\\tan^{-1}x}{2} + C $$<\/p>\n\n\n\n<p>I hope you like article on Integration of tan inverse x interesting and useful. Please do provide the feedback<\/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>Integration of tan inverse x can be calculated using integration by parts .Here is the formula for it \\[ \\int \\tan^{-1}x \\, dx = x \\tan^{-1}x &#8211; \\frac {1}{2} \\ln (1 + x^2) + C \\] where (C) is the constant of integration. Proof of integration of tan inverse x To find the integral we [&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-8767","post","type-post","status-publish","format-standard","hentry","category-maths","tag-integration"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v27.3 - 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Proof of integration of tan inverse x To find the integral we&hellip;","_links":{"self":[{"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/posts\/8767","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=8767"}],"version-history":[{"count":7,"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/posts\/8767\/revisions"}],"predecessor-version":[{"id":8944,"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/posts\/8767\/revisions\/8944"}],"wp:attachment":[{"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/media?parent=8767"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/categories?post=8767"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/tags?post=8767"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}