Indefinite Integration 1 Question 86
87. (i) Show that $\int _0^{\pi} x f(\sin x) d x=\frac{\pi}{2} \int _0^{\pi} f(\sin x) d x$.
(ii) Find the value of $\int _{-1}^{3 / 2}|x \sin \pi x| d x$.
$(1982,2 M)$
$(1982,3 M)$
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Answer:
Correct Answer: 87. (ii) $\frac{3 \pi+1}{\pi^{2}}$
Solution:
- (i) Let $I=\int _0^{\pi} x f(\sin x) d x$
$\Rightarrow \quad I=\int _0^{\pi}(\pi-x) f(\sin x) d x$
On adding Eqs. (i) and (ii), we get
$$ 2 I=\int _0^{\pi} \pi f(\sin x) d x $$
$\therefore \int _0^{\pi} x f(\sin x) d x=\frac{\pi}{2} \int _0^{\pi} f(\sin x) d x$
(ii) Let $\quad I=\int _{-1}^{3 / 2}|x \sin \pi x| d x$
Since, $|x \sin \pi x|=\begin{array}{cc}x \sin \pi x, & -1<x \leq 1 \ -x \sin \pi x, & 1<x<\frac{3}{2}\end{array}$
$\therefore \quad I=\int _{-1}^{1} x \sin \pi x d x+\int _1^{3 / 2}-x \sin \pi x d x$
$=2-\frac{x \cos \pi x}{\pi}{ } _0^{1}-2 \int _0^{1} 1 \cdot \frac{-\cos \pi x}{\pi} d x$
$-\frac{-x \cos \pi x}{\pi}{ } _1^{3 / 2}-\int _1^{3 / 2} \frac{-\cos \pi x}{\pi} d x$
$=\frac{2}{\pi}+\frac{2}{\pi^{2}}(0-0)+\frac{1}{\pi}+\frac{1}{\pi^{2}}(+1-0)$
$=\frac{3}{\pi}+\frac{1}{\pi^{2}}=\frac{3 \pi+1}{\pi^{2}}$
