The value of the integral $\int_{-\pi / 2}^{\pi / 2}\left(x^{2}+\ln \frac{\pi+x}{\pi-x}\right) \cos x \mathrm{~d} x$ is

JEE Advanced · 2012 · 08 Apr 2012 · Paper 2 · Definite Integrals

The value of the integral

\int_{- π /2}^{π /2} (x^{2} + ln \frac{π + x}{π - x}) cos x d x

Answer: (B)

Split the integral:

I = \int_{- π /2}^{π /2} x^{2} cos x d x + \int_{- π /2}^{π /2} ln \frac{π + x}{π - x} cos x d x

The first integrand

x^{2} cos x

is even, so:

\int_{- π /2}^{π /2} x^{2} cos x d x = 2 \int_{0}^{π /2} x^{2} cos x d x

The second integrand

ln \frac{π + x}{π - x} cos x

is odd (since

ln \frac{π + x}{π - x}

is odd and

cos x

is even), so its integral is 0.

Using integration by parts twice on

\int x^{2} cos x d x

\int x^{2} cos x d x = x^{2} sin x + 2 x cos x - 2 sin x + C

Evaluating from 0 to

π /2

[x^{2} sin x + 2 x cos x - 2 sin x]_{0}^{π /2} = \frac{π ^{2}}{4} + 0 - 2 = \frac{π ^{2}}{4} - 2

Therefore:

I = 2 (\frac{π ^{2}}{4} - 2) = \frac{π ^{2}}{2} - 4

JEE Advanced · 2012 · 08 Apr 2012 · Paper 2 · Single correct option