Question

The integral $\int x \,cos^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right) \,dx\, \left(x > 0\right)$ is equal to :

• A. $- x + (1 + x^2) tan^{-1} x + c$
• B. $x- (1 + x^2) cot^{-1} x + c$
• C. $- x + (1 + x^2) cot^{-1} x + c$
• D. $x- (1 + x^2) tan^{-1} x + c$

Answer 1

Answer: (A)

$- x + (1 + x^2) tan^{-1} x + c$

Answer 2

put $\,x = tan\,\theta cos^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right) = cos^{-1}\left(cos2\theta\right) = 2\theta$
$\int tan\,\theta \left(2\theta \right) sec^{2} \,\theta\, d\theta$
$= 2\theta.\int tan\,\theta \, sec^{2} \,\theta \, d\theta - 2\int\left(\frac{d\theta}{d\theta}\int tan\,\theta \, sec^{2} \,\theta \, d\theta \right)d\theta$
$= 2\theta \frac{tan^{2}\theta}{2}-2\int\frac{tan^{2}\theta }{2}d\theta$
$= \theta\,tan^{2}\,\theta-\int\left(sec^{2}\,\theta-1\right)d\theta$
$= \theta \,tan^{2}\,\theta -tan\theta +\theta +C$
$= tan^{-1}x. x^{2}-x + tan^{-1}+C$
$= - x + \left(1 + x^{2}\right) tan^{-1}x + C$