{"id":8937,"date":"2024-01-29T00:02:50","date_gmt":"2024-01-28T18:32:50","guid":{"rendered":"https:\/\/physicscatalyst.com\/article\/?p=8937"},"modified":"2024-01-29T00:04:06","modified_gmt":"2024-01-28T18:34:06","slug":"integration-of-sec-inverse-x","status":"publish","type":"post","link":"https:\/\/physicscatalyst.com\/article\/integration-of-sec-inverse-x\/","title":{"rendered":"integration of sec inverse x"},"content":{"rendered":"\n<p>Integration of sec inverse x can be calculated using integration by parts ,integration by substitution .Here is the formula for it<\/p>\n\n\n\n<p>\\[ \\int \\sec^{-1}x \\, dx = x \\sec^{-1}x &#8211; \\ln |x + \\sqrt { x^2- 1}| + 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 sec inverse x<\/h2>\n\n\n\n<p>We can prove it using integration by parts  and integration by substitution<\/p>\n\n\n\n<p><strong>Integration by parts<\/strong><\/p>\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) = \\sec ^{-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}{x\\sqrt {x^2-1}} \\, 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>$$<br>\\int \\sec^{-1}x \\, dx  = x \\sec^{-1}x  &#8211; \\int x \\cdot \\frac{1}{x\\sqrt {x^2-1}}  \\, dx<br>$$<\/p>\n\n\n\n<p>$ =x \\sec^{-1}x  &#8211; \\int  \\frac{1}{\\sqrt {x^2-1}}  \\, dx $<\/p>\n\n\n\n<p>Now lets calculate the second integral separately  $\\int  \\frac{1}{\\sqrt {x^2 -1}}  \\, dx $<\/p>\n\n\n\n<p>We will use the formula<\/p>\n\n\n\n<p>$\\int \\frac {1}{\\sqrt {x^2 &#8211; a^2}} dx = ln |x + \\sqrt {x^2 &#8211; a^2}| + C$<br><strong>Proof<\/strong><br>Put $x =a sec \\theta$ then $dx= a sec \\theta tan \\theta d\\theta $<br>$\\int \\frac {1}{\\sqrt {x^2 -a^2}} dx $<br>$=\\int sec \\theta d\\theta $<br>$=ln |sec \\theta + tan \\theta| + C $<br>Now<br>$sec \\theta = \\frac {x}{a} $<br>$tan \\theta = \\sqrt {sec^2 \\theta -1 } = \\sqrt {\\frac {x^2}{a^2} -1} $<br>Therefore<br>$=ln | \\frac {x}{a} + \\sqrt {\\frac {x^2}{a^2} -1}| + C_1 = ln |x + \\sqrt {x^2 &#8211; a^2}| &#8211; ln |a| +C_1 = ln |x + \\sqrt {a^2 + x^2}| + C$<\/p>\n\n\n\n<p>Substituting this value in main integral , we get<\/p>\n\n\n\n<p>$$<br>\\int \\sec^{-1}x \\, dx  = x \\sec^{-1}x   &#8211;  \\ln |x + \\sqrt { x^2+ 1}| + C<br>$$<\/p>\n\n\n\n<p><strong>Integration by substitution<\/strong><\/p>\n\n\n\n<p>let $\\theta =  \\sec^{-1}x$<br>$ \\sec \\theta = x$<br>$\\sec \\theta  tan \\theta d \\theta = dx$<\/p>\n\n\n\n<p>$$<br>\\int \\sec^{-1}x \\, dx  = \\int \\theta (\\sec \\theta  tan \\theta) d \\theta<br>$$<\/p>\n\n\n\n<p>Now solving this using integration parts<\/p>\n\n\n\n<p>$ =  \\theta \\int (\\sec \\theta  tan \\theta)  d \\theta  &#8211; \\int (\\frac {d}{d \\theta } \\theta ) .(\\int  (\\sec \\theta tan \\theta)) \\; \\theta \\\\<br>    = \\theta \\sec \\theta &#8211; \\int \\sec \\theta \\; d \\theta $<\/p>\n\n\n\n<p>Now<\/p>\n\n\n\n<p>\\[<br>\\int \\sec(x) \\, dx = \\ln | \\sec(x) + \\tan(x) | + C<br>\\]<\/p>\n\n\n\n<p>Therefore<\/p>\n\n\n\n<p>$=\\theta \\sec \\theta  &#8211;  \\ln | \\sec(x) + \\tan(x) | + C$<\/p>\n\n\n\n<p>Now $tan \\theta= \\sqrt {\\sec^2 \\theta -1} = \\sqrt {x^2-1}$<\/p>\n\n\n\n<p>Substituting back the values<\/p>\n\n\n\n<p>$\\int \\sec^{-1}x \\, dx  = x \\sec^{-1}x   &#8211; \\ln |x +  \\sqrt {x^2 -1}|+ C$<br><\/p>\n\n\n\n<h2 class=\"wp-block-heading\">Solved Examples on  integration of sec inverse x<\/h2>\n\n\n\n<p><strong>Question <\/strong>1<\/p>\n\n\n\n<p>integral of $x \\sec^{-1}x $<\/p>\n\n\n\n<p><strong>Solution<\/strong><\/p>\n\n\n\n<p>Applying integration by parts<br>$\\int u \\, dv = uv &#8211; \\int v \\, du$.<br>Let $u = \\sec^{-1}x$. Then, $du = \\frac{1dx}{x\\sqrt{ x^2-1}}$.<br>Let $dv = x \\, dx$. Then, $v = \\frac{x^2}{2}$.<br>Then<br>$ \\int x \\sec^{-1}x \\, dx = \\frac{x^2}{2} \\sec^{-1}x &#8211; \\int \\frac{x^2}{2} \\cdot \\frac{1dx}{x\\sqrt{x^2 -1}} \\<br>=\\frac{x^2}{2} \\sec^{-1}x &#8211; \\frac {1}{2} \\int \\frac{x}{\\sqrt {x^2-1}} \\; dx$<br>Now lets  solve the integral separately<br> $\\int \\frac{x}{\\sqrt{x^2 -1}} \\; dx$<\/p>\n\n\n\n<p>Let $x^2 -1 = t$<br>$2x dx= dt$<br>therefore<br>$\\int \\frac{x}{\\sqrt{ x^2 -1}} \\; dx =\\frac {1}{2} \\int  t^{-1\/2} \\; dt$<br>$= t^{1\/2} = \\sqrt{x^2 -1}$<\/p>\n\n\n\n<p>Applying these steps to the integral of $x \\sec^{-1}x$, we obtained:<\/p>\n\n\n\n<p>$$\\frac{x^2}{2} \\sec^{-1}x &#8211; \\frac {\\sqrt{x^2 -1}}{2}+ C $$<\/p>\n\n\n\n<p>where $C$ is the constant of integration.<\/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 sec inverse x can be calculated using integration by parts ,integration by substitution .Here is the formula for it \\[ \\int \\sec^{-1}x \\, dx = x \\sec^{-1}x &#8211; \\ln |x + \\sqrt { x^2- 1}| + C \\] where (C) is the constant of integration. Proof of integration of sec inverse x 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":[],"class_list":["post-8937","post","type-post","status-publish","format-standard","hentry","category-maths"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v27.4 - https:\/\/yoast.com\/product\/yoast-seo-wordpress\/ -->\n<title>integration of sec inverse x - 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-sec-inverse-x\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"integration of sec inverse x - physicscatalyst&#039;s Blog\" \/>\n<meta property=\"og:description\" content=\"Integration of sec inverse x can be calculated using integration by parts ,integration by substitution .Here is the formula for it [ int sec^{-1}x , dx = x sec^{-1}x &#8211; ln |x + sqrt { x^2- 1}| + C ] where (C) is the constant of integration. 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Proof of integration of sec inverse x We&hellip;","_links":{"self":[{"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/posts\/8937","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=8937"}],"version-history":[{"count":2,"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/posts\/8937\/revisions"}],"predecessor-version":[{"id":8940,"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/posts\/8937\/revisions\/8940"}],"wp:attachment":[{"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/media?parent=8937"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/categories?post=8937"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/tags?post=8937"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}