Integration of cot inverse x can be calculated using integration by parts .Here is the formula for it<\/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
where (C) is the constant of integration.<\/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
To find the definite integral of $\\cot^{-1}x$ over a specific interval, we use the same approach as with the indefinite integral, but we’ll apply the limits of integration at the end.<\/p>\n\n\n\n
The definite integral of $\\cot^{-1}x$ from $a$ to $b$ is given by:<\/p>\n\n\n\n
$$\\int_{a}^{b} \\cot^{-1}x \\, dx = b \\cot^{-1}b – a \\cot^{-1}a + \\frac {1}{2}\\ln (1 + b^2) – \\frac {1}{2} \\ln (1 + a^2) $$<\/p>\n\n\n\n
This expression represents the accumulated area under the curve of $\\cot^{-1}x$ from $x = a$ to $x = b$.<\/p>\n\n\n\n
Question 1<\/strong><\/p>\n\n\n\n $$\\int_{0}^{1} \\cot^{-1}x \\, dx$$<\/p>\n\n\n\n Solution<\/strong><\/p>\n\n\n\n $\\int_{0}^{1} \\cot^{-1}x \\, dx = [x \\cot^{-1}x + \\frac {1}{2} \\ln (1 + x^2)] _0 ^1 \\\\ Question 2<\/strong><\/p>\n\n\n\n $$ \\int x cot^{-1} x \\, dx$$<\/p>\n\n\n\n Solution<\/strong><\/p>\n\n\n\n Applying integration by parts Applying these steps to the integral of $x \\tan^{-1}x$, we obtained:<\/p>\n\n\n\n $$ \\frac{x^2 \\cot^{-1}x}{2} + \\frac{x }{2} – \\frac{\\tan^{-1}x}{2} + C $$<\/p>\n\n\n\n I hope you like article on Integration of cot inverse x interesting and useful. Please do provide the feedback<\/p>\n\n\n\n Other Integration Related Articles<\/p>\n\n\n\n Integration of cot inverse x can be calculated using integration by parts .Here is the formula for it \\[ \\int \\cot^{-1}x \\, dx = x \\cot^{-1}x + \\frac {1}{2} \\ln (1 + x^2) + C \\] where (C) is the constant of integration. Proof of integration of cot inverse x To find the integral we […]<\/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-8942","post","type-post","status-publish","format-standard","hentry","category-maths"],"yoast_head":"\n
= 1. cot^{-1} 1+ \\frac {\\ln 2}{2} – 0 – \\frac {\\ln 1}{2} = \\frac {\\pi}{4} + \\ln 2\/2$<\/p>\n\n\n\n
$\\int u \\, dv = uv – \\int v \\, du$.
Let $u = \\cot^{-1}x$. Then, $du = \\frac{-dx}{1 + x^2}$.
Let $dv = x \\, dx$. Then, $v = \\frac{x^2}{2}$.
Then
$ \\int x \\cot^{-1}x \\, dx = \\frac{x^2}{2} \\cot^{-1}x + \\int \\frac{x^2}{2} \\cdot \\frac{dx}{1 + x^2} \\
=\\frac{x^2}{2} \\cot^{-1}x + \\frac {1}{2} \\int \\frac{x^2}{1 + x^2} \\; dx$
Now
$\\int \\frac{x^2}{1 + x^2} \\; dx = \\int \\frac{1+x^2 – 1}{1 + x^2} \\; dx \\\\
= \\int 1 dx – \\int \\frac {1}{1+x^2} dx \\\\
= x – tan^{-}x$<\/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":"