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  4. If 2 ∫10 tan-1 xdx = ∫10 cot-1(1 - x + x2)dx, then ∫10 tan-1 (1 - x + x2) dx is equal to :

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Q. If $2 \int^1_0 \tan^{-1} xdx = \int^1_0 \cot^{-1}(1 - x + x^2)dx, $ then $\int^1_0 \tan^{-1} (1 - x + x^2) dx$ is equal to :

JEE MainJEE Main 2016Integrals

Solution:

$2\int\limits^{{1}}_{{0}}tan^{-1}x\,dx=$$\int\limits^{{1}}_{{0}}$ $\left(\frac{\pi}{2}-tan^{-1}\left(1-x+x^{2}\right)\right)dx$
$2\int\limits^{{1}}_{{0}}tan^{-1}x\,dx=$$\int\limits^{{1}}_{{0}}$$\frac{\pi}{2}dx-$$2\int\limits^{{1}}_{{0}}tan^{-1}(1-x+x^2)dx$
$2\int\limits^{{1}}_{{0}}tan^{-1}(1-x+x^2)dx=$$\frac{\pi}{2}-2$$\int\limits^{{1}}_{{0}}tan^{-1}x\,dx$
Let
$I_1 =\int\limits^{{1}}_{{0}}tan^{-1}x\,dx$
$=\left[\left(tan^{-1}x\right)x\right]^{^1}_{0}-$ $\int\limits^{{1}}_{{0}}$$\frac{1}{1+x^{2}}x\,dx$
$=\frac{\pi}{4}-$$\int\limits^{{1}}_{{0}}$$\frac{x}{1+x^{2}}dx$
$=\frac{\pi}{4}-\frac{1}{2}In\,2$
By equation____(1)
$\frac{\pi}{2}-2\left[\frac{\pi}{4}-\frac{1}{2}In\right]=In2$