Indefinite Integration 1 Question 21
22. The value of the integral $\int _{-\pi / 2}^{\pi / 2} x^{2}+\log \frac{\pi-x}{\pi+x} \cos x d x$ is
(a) 0
(b) $\frac{\pi^{2}}{2}-4$
(c) $\frac{\pi^{2}}{2}+4$
(d) $\frac{\pi^{2}}{2}$
(2012)
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Solution:
- $I=\int _{-\pi / 2}^{\pi / 2} x^{2}+\log \frac{\pi-x}{\pi+x} \cos x d x$
As, $\quad \int _{-a}^{a} f(x) d x=0$, when $f(-x)=-f(x)$
$$ \begin{aligned} \therefore \quad I & =\int _{-\pi / 2}^{\pi / 2} x^{2} \cos x d x+0=2 \int _0^{\pi / 2}\left(x^{2} \cos x\right) d x \\ & =2{\left(x^{2} \sin x\right) _0^{\pi / 2}-\int _0^{\pi / 2} 2 x \cdot \sin x d x } \\ & =2 \frac{\pi^{2}}{4}-2{(-x \cdot \cos x) _0^{\pi / 2}-\int _0^{\pi / 2} 1 \cdot(-\cos x) d x } \\ & =2 \frac{\pi^{2}}{4}-2(\sin x) _0^{\pi / 2}=2 \frac{\pi^{2}}{4}-2=\frac{\pi^{2}}{2}-4 \end{aligned} $$
