{"id":8628,"date":"2024-01-05T18:03:39","date_gmt":"2024-01-05T12:33:39","guid":{"rendered":"https:\/\/physicscatalyst.com\/article\/?p=8628"},"modified":"2024-01-17T23:19:48","modified_gmt":"2024-01-17T17:49:48","slug":"integration-of-sec-x","status":"publish","type":"post","link":"https:\/\/physicscatalyst.com\/article\/integration-of-sec-x\/","title":{"rendered":"Integration of sec x"},"content":{"rendered":"\n

Integration of sec x can be found using various integration technique like integration by substitution, Integration by partial fraction along with trigonometric identities. The various formula for integration of sec x are<\/p>\n\n\n\n

I<\/p>\n\n\n\n

\\[
\\int \\sec(x) \\, dx = \\ln | \\sec(x) + \\tan(x) | + C
\\]<\/p>\n\n\n\n

II<\/p>\n\n\n\n

\\[
\\int \\sec(x) \\, dx = -\\ln | \\sec(x) – \\tan(x) | + C
\\]<\/p>\n\n\n\n

III<\/p>\n\n\n\n

\\[
\\int \\csc(x) \\, dx = \\ln | \\tan ( \\frac {x}{2} + \\frac {\\pi}{2})| + C
\\]<\/p>\n\n\n\n

IV<\/p>\n\n\n\n

\\[
\\int \\csc(x) \\, dx = \\frac {1}{2} \\ln | \\frac {1 + \\sin x}{1 – \\sin x} | + C
\\]<\/p>\n\n\n\n

Lets check out the proof of each of these<\/p>\n\n\n\n

Proof of I<\/h2>\n\n\n\n

Integral of secant can be little tricky and can be solved by multiplying and dividing by $(\\sec(x) + \\tan(x))$. Here’s how it’s done:<\/p>\n\n\n\n

\\[
\\int \\sec(x) \\, dx = \\int \\sec(x) \\cdot \\frac{\\sec(x) + \\tan(x)}{\\sec(x) + \\tan(x)} \\, dx \\\\
= \\int \\frac{\\sec^2(x) + \\sec(x)\\tan(x)}{\\sec(x) + \\tan(x)} \\, dx.
\\]<\/p>\n\n\n\n

Now, taking u as
$ u= \\sec(x) + \\tan(x) $
$ du =(\\sec(x)\\tan(x) + \\sec^2(x)) dx $<\/p>\n\n\n\n

Therefore integration becomes<\/p>\n\n\n\n

\\[
\\int \\sec(x) \\, dx =\\int \\frac {1}{u} \\, du,
\\]<\/p>\n\n\n\n

hence<\/p>\n\n\n\n

\\[
\\int \\sec(x) \\, dx = \\ln | \\sec(x) + \\tan(x) | + C
\\]<\/p>\n\n\n\n

Proof of II<\/h2>\n\n\n\n

We can write the result of I as<\/p>\n\n\n\n

\\[
\\int \\sec(x) \\, dx = \\ln | \\sec(x) + \\tan(x) | + C = \\ln | \\frac {\\sec^2(x) – \\tan^2(x)}{\\sin(x) – \\sin(x) } | + C \\\\
=\\ln | \\frac {1}{\\sec(x) – \\tan(x) } | + C = \\ln | (\\sec(x) – \\tan(x))^{-1} | + C \\\\
=- \\ln | \\sec(x) – \\tan(x) | + C
\\]<\/p>\n\n\n\n

Proof of III<\/h2>\n\n\n\n

This can be provided using trigonometric formulas<\/p>\n\n\n\n

\\[
\\int \\sec(x) \\, dx =\\int \\frac {1}{\\cos(x)} \\, dx
=\\int \\frac {1}{\\sin(x + \\frac {\\pi}{2})} \\, dx
=\\int \\frac {1}{2 \\sin(x\/2 + \\pi\/4) \\cos(x\/2 + \\pi\/4)} \\, dx
\\]<\/p>\n\n\n\n

Dividing and Multiplying $cos(x\/2 + \\pi\/4)$ in the denominator<\/p>\n\n\n\n

\\[
\\int \\sec(x) \\, dx = \\frac {1}{2}\\int \\frac {\\sec^2(x\/2+ \\pi\/4)}{\\tan(x\/2+ \\pi\/4)} \\, dx
\\]<\/p>\n\n\n\n

Lets $u=\\tan(x\/2 + \\pi\/4)$
$du= \\frac{1}{2} \\sec^2(x\/2 + \\pi\/4) dx$<\/p>\n\n\n\n

Therefore<\/p>\n\n\n\n

\\[
\\int \\sec(x) \\, dx = \\int \\frac {1}{u} \\, du = ln |u| + C
\\]<\/p>\n\n\n\n

Hence<\/p>\n\n\n\n

\\[
\\int \\sec(x) \\, dx = \\ln | \\tan (\\frac {x}{2} +\\frac {\\pi}{2}) | + C
\\]<\/p>\n\n\n\n

Proof of IV<\/h2>\n\n\n\n

This can be proof by using integration by partial fractions<\/p>\n\n\n\n

\\[
\\int \\sec(x) \\, dx =\\int \\frac {1}{\\cos(x)} \\, dx
=\\int \\frac {cos(x)}{\\cos^2(x)} \\, dx
=\\int \\frac {cos(x)}{ 1- \\sin^2(x)} \\, dx
\\]<\/p>\n\n\n\n

Lets $ u=\\sin(x)$
$du= \\cos(x) dx$<\/p>\n\n\n\n

Therefore<\/p>\n\n\n\n

\\[
\\int \\sec(x) \\, dx =\\int \\frac {1}{1 -u^2} \\, du
=\\frac {1}{2} \\int [\\frac {1}{1+u} + \\frac {1}{1-u} ]\\, du
=\\frac {1}{2} ln |\\frac {1+u}{1-u}| + C
\\]<\/p>\n\n\n\n

Hence<\/p>\n\n\n\n

\\[
\\int \\csc(x) \\, dx = \\frac {1}{2} \\ln | \\frac {1 + \\sin x}{ 1- \\sin x} | + C
\\]<\/p>\n\n\n\n

Solved Example<\/h2>\n\n\n\n

Example 1<\/strong><\/p>\n\n\n\n

\\[
\\int_{0}^{\\pi\/4} \\sec(x) \\, dx.
\\]<\/p>\n\n\n\n

Solution<\/strong><\/p>\n\n\n\n

Using the result from above:<\/p>\n\n\n\n

\\[
\\int_{0}^{\\pi\/4} \\sec(x) \\, dx = \\left[ \\ln|\\sec(x) + \\tan(x)| \\right]_{0}^{\\pi\/4}.
\\]<\/p>\n\n\n\n

Evaluating at the bounds:<\/p>\n\n\n\n

\\[
\\begin{align} \\left[ \\ln|\\sec(x) + \\tan(x)| \\right]_{0}^{\\pi\/4}
&= \\ln|\\sec(\\pi\/4) + \\tan(\\pi\/4)| – \\ln|\\sec(0) + \\tan(0)| \\\\
&= \\ln|(\\sqrt{2}) + 1| – \\ln|1 + 0| \\\\
&= \\ln(\\sqrt{2} + 1). \\end{align}
\\]<\/p>\n\n\n\n

So, $\\int_{0}^{\\pi\/4} \\sec(x) \\, dx = \\ln(\\sqrt{2} + 1)$.<\/p>\n\n\n\n

\n

Other Integration Related Articles<\/p>\n\n\n\n

Integration of tan inverse x<\/a><\/td>Integration of cos inverse x<\/a><\/td><\/tr>
Integration of cos square x<\/a><\/td>Integration of fractional part of x<\/a><\/td><\/tr>
integration of uv formula<\/a><\/td>integration of log x<\/a><\/td><\/tr>
integration of sin inverse x<\/a><\/td>Integration of constant term<\/a><\/td><\/tr>
integration of hyperbolic functions<\/a><\/td>List of Integration Formulas <\/a><\/td><\/tr>
integration of log sinx<\/a><\/td>integration of sin x cos x dx<\/a><\/td><\/tr><\/tbody><\/table><\/figure>\n<\/div><\/div>\n","protected":false},"excerpt":{"rendered":"

Integration of sec x can be found using various integration technique like integration by substitution, Integration by partial fraction along with trigonometric identities. The various formula for integration of sec x are I \\[\\int \\sec(x) \\, dx = \\ln | \\sec(x) + \\tan(x) | + C\\] II \\[\\int \\sec(x) \\, dx = -\\ln | \\sec(x) […]<\/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-8628","post","type-post","status-publish","format-standard","hentry","category-maths","tag-integration"],"yoast_head":"\nIntegration of sec x - physicscatalyst'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-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 x - physicscatalyst's Blog\" \/>\n<meta property=\"og:description\" content=\"Integration of sec x can be found using various integration technique like integration by substitution, Integration by partial fraction along with trigonometric identities. 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The various formula for integration of sec x are I \\[\\int \\sec(x) \\, dx = \\ln | \\sec(x) + \\tan(x) | + C\\] II \\[\\int \\sec(x) \\, dx = -\\ln | \\sec(x)…","_links":{"self":[{"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/posts\/8628","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=8628"}],"version-history":[{"count":2,"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/posts\/8628\/revisions"}],"predecessor-version":[{"id":8806,"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/posts\/8628\/revisions\/8806"}],"wp:attachment":[{"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/media?parent=8628"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/categories?post=8628"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/tags?post=8628"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}