Integration of inverse trigonometric functions can be calculated using integration by parts ,integration by substitution .Here are the formula for it<\/p>\n\n\n\n
\\[ \\int \\sin^{-1}x \\, dx = x \\sin^{-1}x + \\sqrt{1 – x^2} + C \\]<\/p>\n\n\n\n
\\[ \\int \\cos^{-1}x \\, dx = x \\cos^{-1}x – \\sqrt{1 – x^2} + C \\]<\/p>\n\n\n\n
\\[ \\int \\tan^{-1}x \\, dx = x \\tan^{-1}x – \\frac {1}{2} \\ln (1 + x^2) + C \\]<\/p>\n\n\n\n
\\[ \\int \\csc^{-1}x \\, dx = x \\csc^{-1}x + \\ln |x + \\sqrt { x^2+ 1}| + C \\]<\/p>\n\n\n\n
\\[ \\int \\sec^{-1}x \\, dx = x \\sec^{-1}x – \\ln |x + \\sqrt { x^2- 1}| + C \\]<\/p>\n\n\n\n
\\[ \\int \\cot^{-1}x \\, dx = x \\cot^{-1}x + \\frac {1}{2} \\ln (1 + x^2) + C \\]<\/p>\n\n\n\n
let $\\theta = \\sin^{-1}x$
$ sin \\theta = x$
$cos \\theta d \\theta = dx$<\/p>\n\n\n\n
$$
\\int \\sin^{-1}x \\, dx = \\int \\theta cos \\theta d \\theta
$$<\/p>\n\n\n\n
Now solving this using integration parts<\/p>\n\n\n\n
$ = \\theta \\int cos \\theta d \\theta – \\int (\\frac {d}{d \\theta } \\theta ) .(\\int cos \\theta d \\theta) \\; \\theta \\\\
= \\theta \\sin \\theta – \\int \\sin \\theta \\; d \\theta \\\\
=\\theta \\sin \\theta + \\cos \\theta $<\/p>\n\n\n\n
Substituting back the values<\/p>\n\n\n\n
$\\int \\sin^{-1}x \\, dx = x \\sin^{-1}x + \\sqrt {1-x^2} + C$<\/p>\n\n\n\n
let $\\theta = \\cos^{-1}x$
$ cos \\theta = x$
$-sin \\theta d \\theta = dx$<\/p>\n\n\n\n
$$
\\int \\cos^{-1}x \\, dx = -\\int \\theta sin \\theta d \\theta
$$<\/p>\n\n\n\n
Now solving this using integration parts<\/p>\n\n\n\n
$ = -\\theta \\int sin \\theta d \\theta + \\int (\\frac {d}{d \\theta } \\theta ) .(\\int sin \\theta d \\theta) \\; \\theta \\\\
= \\theta \\cos \\theta – \\int \\cos \\theta \\; d \\theta \\\\
=\\theta \\cos \\theta – \\sin \\theta $<\/p>\n\n\n\n
Substituting back the values<\/p>\n\n\n\n
$\\int \\cos^{-1}x \\, dx = x \\cos^{-1}x – \\sqrt {1-x^2} + C$<\/p>\n\n\n\n
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
$\\int f(x) g(x) dx = f(x) (\\int g(x) dx )- \\int \\left \\{ \\frac {df(x)}{dx} \\int g(x) dx \\right \\} dx $
In our case, we can let $ f(x) = \\tan^{-1}x $ and $ g(x) = 1 $. Then<\/p>\n\n\n\n
Now, substitute these into the integration by parts formula:<\/p>\n\n\n\n
$$
\\int \\tan^{-1}x \\, dx = x \\tan^{-1}x – \\int x \\cdot \\frac{1}{1+x^2} \\, dx
$$<\/p>\n\n\n\n
$ =x \\tan^{-1}x – \\int \\frac{x}{1+x^2} \\, dx \\\\
=x \\tan^{-1}x – \\frac {1}{2} \\int \\frac{2x}{1+x^2} \\, dx $<\/p>\n\n\n\n
Now lets calculate the second integral separately $\\int \\frac{2x}{1+x^2} \\, dx $<\/p>\n\n\n\n
Let $t= 1+x^2$
then $dt=2x dx$
Therefore<\/p>\n\n\n\n
$\\int \\frac{2x}{1+x^2} \\, dx = \\int \\frac{1}{t} \\, dx \\\\
= \\ln |t| = \\ln |1+x^2|$<\/p>\n\n\n\n
Substituting this value in main integral , we get<\/p>\n\n\n\n
$$
\\int \\tan^{-1}x \\, dx = x \\tan^{-1}x – \\frac {1}{2} \\ln (1 + x^2) + C $$<\/p>\n\n\n\n
let $\\theta = \\csc^{-1}x$
$ \\csc \\theta = x$
$-\\csc \\theta cot \\theta d \\theta = dx$<\/p>\n\n\n\n
$$
\\int \\csc^{-1}x \\, dx = \\int \\theta (-\\csc \\theta cot \\theta) d \\theta
$$<\/p>\n\n\n\n
Now solving this using integration parts<\/p>\n\n\n\n
$ = \\theta \\int (-\\csc \\theta cot \\theta) d \\theta – \\int (\\frac {d}{d \\theta } \\theta ) .(\\int (-\\csc \\theta cot \\theta)) \\; \\theta \\\\
= \\theta \\csc \\theta – \\int \\csc \\theta \\; d \\theta $<\/p>\n\n\n\n
Now<\/p>\n\n\n\n
\\[
\\int \\csc(x) \\, dx = \\ln | \\csc(x) – \\cot(x) | + C
\\]<\/p>\n\n\n\n
Therefore<\/p>\n\n\n\n
$=\\theta \\csc \\theta – \\ln | \\csc(x) – \\cot(x) | + C$<\/p>\n\n\n\n
Now $cot \\theta= \\sqrt {\\csc^2 \\theta -1} = \\sqrt {x^2-1}$<\/p>\n\n\n\n
Substituting back the values<\/p>\n\n\n\n
$\\int \\csc^{-1}x \\, dx = x \\csc^{-1}x – \\ln |x – \\sqrt {x^2 -1}|+ C$
$= x \\csc^{-1}x + \\ln |\\frac {1}{x – \\sqrt {x^2 -1}}| + C$
$= x \\csc^{-1}x + \\ln |x + \\sqrt {x^2 -1}| + C$<\/p>\n\n\n\n
let $\\theta = \\sec^{-1}x$
$ \\sec \\theta = x$
$\\sec \\theta tan \\theta d \\theta = dx$<\/p>\n\n\n\n
$$
\\int \\sec^{-1}x \\, dx = \\int \\theta (\\sec \\theta tan \\theta) d \\theta
$$<\/p>\n\n\n\n
Now solving this using integration parts<\/p>\n\n\n\n
$ = \\theta \\int (\\sec \\theta tan \\theta) d \\theta – \\int (\\frac {d}{d \\theta } \\theta ) .(\\int (\\sec \\theta tan \\theta)) \\; \\theta \\\\
= \\theta \\sec \\theta – \\int \\sec \\theta \\; d \\theta $<\/p>\n\n\n\n
Now<\/p>\n\n\n\n
\\[
\\int \\sec(x) \\, dx = \\ln | \\sec(x) + \\tan(x) | + C
\\]<\/p>\n\n\n\n
Therefore<\/p>\n\n\n\n
$=\\theta \\sec \\theta – \\ln | \\sec(x) + \\tan(x) | + C$<\/p>\n\n\n\n
Now $tan \\theta= \\sqrt {\\sec^2 \\theta -1} = \\sqrt {x^2-1}$<\/p>\n\n\n\n
Substituting back the values<\/p>\n\n\n\n
$\\int \\sec^{-1}x \\, dx = x \\sec^{-1}x – \\ln |x + \\sqrt {x^2 -1}|+ C$<\/p>\n\n\n\n
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
$\\int f(x) g(x) dx = f(x) (\\int g(x) dx )- \\int \\left \\{ \\frac {df(x)}{dx} \\int g(x) dx \\right \\} dx $
In our case, we can let $ f(x) = \\cot^{-1}x $ and $ g(x) = 1 $. Then<\/p>\n\n\n\n
Now, substitute these into the integration by parts formula:<\/p>\n\n\n\n
$$
\\int \\cot^{-1}x \\, dx = x \\cot^{-1}x – \\int x \\cdot \\frac{-1}{1+x^2} \\, dx
$$<\/p>\n\n\n\n
$ =x \\tan^{-1}x + \\int \\frac{x}{1+x^2} \\, dx \\\\
=x \\tan^{-1}x – \\frac {1}{2} \\int \\frac{2x}{1+x^2} \\, dx $<\/p>\n\n\n\n
Now lets calculate the second integral separately $\\int \\frac{2x}{1+x^2} \\, dx $<\/p>\n\n\n\n
Let $t= 1+x^2$
then $dt=2x dx$
Therefore<\/p>\n\n\n\n
$\\int \\frac{2x}{1+x^2} \\, dx = \\int \\frac{1}{t} \\, dx \\\\
= \\ln |t| = \\ln |1+x^2|$<\/p>\n\n\n\n
Substituting this value in main integral , we get<\/p>\n\n\n\n
$$
\\int \\cot^{-1}x \\, dx = x \\cot^{-1}x + \\frac {1}{2} \\ln (1 + x^2) + C $$<\/p>\n\n\n\n
Integrals that result in inverse trigonometric functions, like $\\int \\frac{dx}{\\sqrt{1 – x^2}} = \\sin^{-1}(x) + C$ or $\\int \\frac{dx}{1 + x^2} = \\tan^{-1}(x) + C$.<\/p>\n\n\n\n
$\\int ( \\frac {1}{\\sqrt {1-x^2} } ) = \\sin^{-1}x + C$<\/p>\n\n\n\n
$\\int (\\frac {1}{\\sqrt {1-x^2}}) = – \\cos ^{-1}x +C$<\/p>\n\n\n\n
$\\int ( \\frac {1}{1 + x^2}) =\\tan ^{-1}x + C$<\/p>\n\n\n\n
$\\int ( \\frac {1}{1 + x^2}) = -\\cot ^{-1}x + C$<\/p>\n\n\n\n
$\\int (\\frac {1}{|x|\\sqrt {x^-1}}) = -sec^{-1} x + C $<\/p>\n\n\n\n
$\\int (\\frac {1}{|x|\\sqrt {x^-1}}) = -cosec^{-1} x + C $<\/p>\n\n\n\n
<\/p>\n\n\n\n
(1)What are inverse trigonometric functions? (2)Why are inverse trigonometric functions important in integration? I hope you like this post on Integration of inverse trigonometric functions. Please do provide the feedback<\/p>\n\n\n\n Other Integration Related Articles<\/p>\n\n\n\n Integration of inverse trigonometric functions can be calculated using integration by parts ,integration by substitution .Here are the formula for it \\[ \\int \\sin^{-1}x \\, dx = x \\sin^{-1}x + \\sqrt{1 – x^2} + C \\] \\[ \\int \\cos^{-1}x \\, dx = x \\cos^{-1}x – \\sqrt{1 – x^2} + C \\] \\[ \\int \\tan^{-1}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":[],"class_list":["post-8976","post","type-post","status-publish","format-standard","hentry","category-maths"],"yoast_head":"\n
Answer<\/strong> Inverse trigonometric functions are functions that reverse the action of the trigonometric functions. They include $sin^{-1}(x)$, $cos^{-1}(x)$, $tan^{-1}(x)$, $cot^{-1}(x)$, $sec^{-1}(x)$, and $csc^{-1}(x)$, and are used to find the angle whose trigonometric function equals a given number.<\/p>\n\n\n\n
Answer<\/strong> They are important because they appear in the solutions of integrals involving trigonometric expressions, especially in cases where direct integration is not straightforward. They help in solving integrals related to geometric and physical problems.<\/p>\n\n\n\nIntegration of modulus function<\/a><\/td> Integration of tan x<\/a><\/td> <\/td> <\/td><\/tr> Integration of sec x<\/a><\/td> Integration of greatest integer function<\/a><\/td> <\/td> <\/td><\/tr> Integration of trigonometric functions<\/a><\/td> Integration of cot x<\/a><\/td> <\/td> <\/td><\/tr> Integration of cosec x<\/a><\/td> Integration of sinx<\/a><\/td> <\/td> <\/td><\/tr> Integration of irrational functions<\/a><\/td> integration of root x<\/a><\/td> <\/td> <\/td><\/tr> Integration of fractional part of x<\/a><\/td> Integration of cos square x<\/a><\/td> <\/td> <\/td><\/tr><\/tbody><\/table><\/figure>\n<\/div><\/div>\n","protected":false},"excerpt":{"rendered":"