The value of the integral ∫_ -π / 2 ^ π / 2 (x^ 2 + (π+x)/(π-x)) x d x is

Correct answer: B. π^ 2 2 -4 ∫_ -π / 2 ^ π / 2 [x^ 2 + ((π+x)/(π-x))] x d x =∫_ -π / 2 ^ π / 2 x^ 2 x d x+∫_ -π / 2 ^ π / 2 ((π+x)/(π-x)) x d x =2 ∫_ 0 ^ π / 2 x^ 2 x d x+0 [ x^ 2 x. is an even function and ((π+x)/(π-x)) x is an odd function] =2[x^ 2 x+2 x x-2 x]_ 0 ^ π / 2 = π^ 2 2 -4

JEE Advanced 2012Paper 2

Exam: JEE Advanced
Year: 2012
Paper: Paper 2
Subject: Mathematics
Chapter: Definite Integration

JEE Advanced 2012 · Paper 2 · Mathematics · Definite Integration

The value of the integral ∫_ -π / 2 ^ π / 2 (x^ 2 + (π+x)/(π-x)) x...

The value of the integral

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

Options

A0
B $\frac{\pi^{2}}{2}-4$
C $\frac{\pi^{2}}{2}+4$
D $\frac{\pi^{2}}{2}$

Correct answer

B. $\frac{\pi^{2}}{2}-4$

Step-by-step solution

\int_{-\pi / 2}^{\pi / 2}\left[x^{2}+\ln \left(\frac{\pi+x}{\pi-x}\right)\right] \cos x d x

=\int_{-\pi / 2}^{\pi / 2} x^{2} \cos x d x+\int_{-\pi / 2}^{\pi / 2} \ln \left(\frac{\pi+x}{\pi-x}\right) \cos x d x

=2 \int_{0}^{\pi / 2} x^{2} \cos x d x+0 \quad\left[\because x^{2} \cos x\right.

is an even function and

\ln \left(\frac{\pi+x}{\pi-x}\right) \cos x

is an odd function]

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

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