JEE Main 2014 (Online) 11th AprilQuestion 1
The integral $\int x\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)dx\ (x>0)$ is equal to :
Indefinite Integration
Solution
Concept. For $x>0$, $\cos^{-1}\frac{1-x^2}{1+x^2}=2\tan^{-1}x$; then integrate by parts. Formulas used. (1) $\cos2\theta=\dfrac{1-\tan^2\theta}{1+\tan^2\theta}$, so with $x=\tan\theta$, $\theta\in\left(0,\frac\pi2\right)$: $\cos^{-1}\dfrac{1-x^2}{1+x^2}=2\theta$(2) $\int u\,dv=uv-\int v\,du$Step 1 - Simplify. By (1), the integrand is $2x\tan^{-1}x$. Step 2 - By parts. With $u=\tan^{-1}x$, $dv=2x\,dx$: $$x^2\tan^{-1}x-\int\frac{x^2}{1+x^2}\,dx=x^2\tan^{-1}x-x+\tan^{-1}x+c.$$ Step 3 - Collect. $-x+(1+x^2)\tan^{-1}x+c$. Answer: (A)