Question

When $x>0,$ men $\int {\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)dx$ is

• A. $2[x{\tan }^{-1}x-\log (1+x^2)]+c$
• B. 2$[x{\tan }^{-1}x+\log (1\left)\right]+c$
• C. $2x{\tan }^{-1}x+\log (1+x^2)+c$
• D. $2x{\tan }^{-1}x-\log (1+x^2)+c$

Answer 1

Answer: (D)

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

$Letx=tan\theta or\theta =ta{n}^{-1}x$

then

$I=\int co{s}^{-1}\left(\right(\frac{1-ta{n}^2\theta }{1+ta{n}^2\theta }\left)\right)dx$

$=\int co{s}^{-1}\left(cos2\theta \right)dx$

$=\int 2\theta dx$

$=2\int ta{n}^{-1}x\times 1dx$

$=2[ta{n}^{-1}(\int 1.dx)-\int (\frac{1}{1+x^2}\left)(\int 1.dx)dx\right]$

$=2[xta{n}^{-1}x-\int (\frac{x}{1+x^2}\left)dx\right]$

$let1+x^2=t$

$then2xdx=dt$

$\therefore I=2[xta{n}^{-1}x-(\frac{1}{2}\left)log\right(1+x^2\left)\right]=C$

$=2xta{n}^{-1}x-log|1+x^2|+C$