IIT-JEE 2012 · Paper 2 · Mathematics Q47

MathematicsSingle correct+3 / -1Question 47 of 60

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

\dfrac{\pi^2}2-4

\dfrac{\pi^2}2+4

\dfrac{\pi^2}2

Answer and worked solution

Official answer: B

Odd part vanishes. $\ln\frac{\pi+x}{\pi-x}$ is odd and $\cos x$ is even, so that part integrates to 0.

Even part.
$\int_{-\pi/2}^{\pi/2}x^2\cos x\,dx=2\Big[x^2\sin x+2x\cos x-2\sin x\Big]_0^{\pi/2}=2\left(\frac{\pi^2}4-2\right)=\frac{\pi^2}2-4.$

(B)

Topics:Properties of definite integrals Integration by parts

IIT-JEE 2012 · Paper 2, Mathematics question 47