Question

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

• A. $-x+(1+x^2)ta{n}^{-1}x+c$
• B. $x-(1+x^2)co{t}^{-1}x+c$
• C. $-x+(1+x^2)co{t}^{-1}x+c$
• D. $x-(1+x^2)ta{n}^{-1}x+c$

Answer 1

Answer: (A)

put $x=tan\theta co{s}^{-1}\left(\frac{1-x^2}{1+x^2}\right)=co{s}^{-1}\left(cos2\theta \right)=2\theta$
$\int tan\theta \left(2\theta \right)se{c}^2\theta d\theta$
$=2\theta .\int tan\theta se{c}^2\theta d\theta -2\int \left(\frac{d\theta }{d\theta }\int tan\theta se{c}^2\theta d\theta \right)d\theta$
$=2\theta \frac{ta{n}^2\theta }{2}-2\int \frac{ta{n}^2\theta }{2}d\theta$
$=\theta ta{n}^2\theta -\int \left(se{c}^2\theta -1\right)d\theta$
$=\theta ta{n}^2\theta -tan\theta +\theta +C$
$=ta{n}^{-1}x.x^2-x+ta{n}^{-1}+C$
$=-x+\left(1+x^2\right)ta{n}^{-1}x+C$