The value of the integral $\int_{0}^{1} x \cot ^{-1}\left(1-x^{2}+x^{4}\right) dx$ is

(A) $\frac{\pi}{4}-\frac{1}{2} \log _{e} 2$

(B) $\frac{\pi}{2}-\log _{e} 2$

(C) $\frac{\pi}{2}-\frac{1}{2} \log _{e} 2$

(D) $\frac{\pi}{4}-\log _{e} 2$

JEE Main · 2019 · 09 Apr 2019 · Shift 2 · Definite Integrals

The value of the integral

\int_{0}^{1} x cot^{- 1} (1 - x^{2} + x^{4}) d x

Answer: (A)

Rewrite

cot^{- 1} (1 - x^{2} + x^{4}) = tan^{- 1} \frac{1}{1 - x ^{2} + x ^{4}}

, and notice

\frac{1}{1 - x ^{2} + x ^{4}} = \frac{x ^{2} - ( x ^{2} - 1 )}{1 + x ^{2} ( x ^{2} - 1 )}

, which is the subtraction formula for

tan^{- 1} (x^{2}) - tan^{- 1} (x^{2} - 1)

. So

I = \int_{0}^{1} x [tan^{- 1} (x^{2}) - tan^{- 1} (x^{2} - 1)] d x

. Substituting

t = x^{2}

with

d t = 2 x d x

turns this into

\frac{1}{2} \int_{0}^{1} [tan^{- 1} t - tan^{- 1} (t - 1)] d t

, and by symmetry over

[0, 1]

it reduces to

\int_{0}^{1} tan^{- 1} t d t = [t tan^{- 1} t]_{0}^{1} - \int_{0}^{1} \frac{t}{1 + t ^{2}} d t = \frac{π}{4} - \frac{1}{2} lo g_{e} 2

JEE Main · 2019 · 09 Apr 2019 · Shift 2 · Single correct option