Definite Integration Question 75
Question 75
- Evaluate $\int_{0}^{\pi} \frac{x \sin (2 x) \sin \frac{\pi}{2} \cos x}{2 x-\pi} d x$.
$(1991,4$ M)
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Answer:
Correct Answer: 77. $\frac{8}{\pi^{2}}$
Solution:
- Let $I=\int_{0}^{\pi} \frac{x \sin (2 x) \cdot \sin \frac{\pi}{2} \cos x}{(2 x-\pi)} d x$
Then $I=\int_{0}^{\pi} \frac{(\pi-x) \cdot \sin 2(\pi-x) \cdot \sin \frac{\pi}{2} \cos (\pi-x)}{2(\pi-x)-\pi} d x$
$$ \begin{aligned} & \Rightarrow \quad I=\int_{0}^{\pi} \frac{(\pi-x) \cdot \sin 2 x \cdot \sin \frac{\pi}{2} \cos x}{\pi-2 x} d x \ & \Rightarrow \quad I=\int_{0}^{\pi} \frac{(x-\pi) \sin 2 x \cdot \sin \frac{\pi}{2} \cos x}{(2 x-\pi)} d x \end{aligned} $$
On adding Eqs. (i) and (iii), we get
$$ \begin{array}{cc} & 2 I=\int_{0}^{\pi} \sin 2 x \cdot \sin \frac{\pi}{2} \cos x d x \ \Rightarrow \quad & 2 I=2 \int_{0}^{\pi} \sin x \cos x \cdot \sin \frac{\pi}{2} \cos x d x \ \Rightarrow \quad & I=\int_{0}^{\pi} \sin x \cos x \cdot \sin \frac{\pi}{2} \cos x d x \end{array} $$
put $\frac{\pi}{2} \cos x=t \Rightarrow-\frac{\pi}{2} \sin x d x=d t \Rightarrow \sin x d x=-\frac{2}{\pi} d t$
$\therefore \quad I=-\frac{2}{\pi} \int_{\pi / 2}^{-\pi / 2} \frac{2 t}{\pi} \cdot \sin t d t$
$=\frac{4}{\pi^{2}} \int_{-\pi / 2}^{\pi / 2} t \sin t d t$
$\Rightarrow \quad I=\frac{4}{\pi^{2}}[-t \cos t+\sin t]_{-\pi / 2}^{\pi / 2}=\frac{4}{\pi^{2}} \times 2=\frac{8}{\pi^{2}}$
