Indefinite Integration 1 Question 76

77. 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:

  1. 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 $\frac{\pi}{2} \cos x=t \Rightarrow-\frac{\pi}{2} \sin x d x=d t \Rightarrow \sin x d x=-\frac{2}{\pi} d t$

$\therefore \quad I=-\frac{2}{\pi} \int _{\pi / 2}^{-\pi / 2} \frac{2 t}{\pi} \cdot \sin t d t$

$=\frac{4}{\pi^{2}} \int _{-\pi / 2}^{\pi / 2} t \sin t d t$

$\Rightarrow \quad I=\frac{4}{\pi^{2}}[-t \cos t+\sin t] _{-\pi / 2}^{\pi / 2}=\frac{4}{\pi^{2}} \times 2=\frac{8}{\pi^{2}}$



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