SATHEE: Indefinite Integration 1 Question 76

Indefinite Integration 1 Question 76

77. Evaluate $\int_{0}^{π} \frac{x \sin (2 x) \sin \frac{π}{2} \cos x}{2 x - π} d x$ .

$(1991, 4$ M)

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Answer:

Correct Answer: 77. $\frac{8}{π^{2}}$

Solution:

Let $I = \int_{0}^{π} \frac{x \sin (2 x) \cdot \sin \frac{π}{2} \cos x}{(2 x - π)} d x$

Then $I = \int_{0}^{π} \frac{(π - x) \cdot \sin 2 (π - x) \cdot \sin \frac{π}{2} \cos (π - x)}{2 (π - x) - π} d x$

$\begin{aligned} \Rightarrow I = \int_{0}^{π} \frac{(π - x) \cdot \sin 2 x \cdot \sin \frac{π}{2} \cos x}{π - 2 x} d x \\ \Rightarrow I = \int_{0}^{π} \frac{(x - π) \sin 2 x \cdot \sin \frac{π}{2} \cos x}{(2 x - π)} d x \end{aligned}$

On adding Eqs. (i) and (iii), we get

$\begin{array}{cc} 2 I = \int_{0}^{π} \sin 2 x \cdot \sin \frac{π}{2} \cos x d x \\ \Rightarrow & 2 I = 2 \int_{0}^{π} \sin x \cos x \cdot \sin \frac{π}{2} \cos x d x \\ \Rightarrow & I = \int_{0}^{π} \sin x \cos x \cdot \sin \frac{π}{2} \cos x d x \end{array}$

put $\frac{π}{2} \cos x = t \Rightarrow - \frac{π}{2} \sin x d x = d t \Rightarrow \sin x d x = - \frac{2}{π} d t$

$∴ I = - \frac{2}{π} \int_{π / 2}^{- π / 2} \frac{2 t}{π} \cdot \sin t d t$

$= \frac{4}{π^{2}} \int_{- π / 2}^{π / 2} t \sin t d t$

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

Indefinite Integration 1 Question 76

77. Evaluate $\int_{0}^{π} \frac{x \sin (2 x) \sin \frac{π}{2} \cos x}{2 x - π} d x$ .

Answer:

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Indefinite Integration 1 Question 76

77. Evaluate ∫0πxsin⁡(2x)sin⁡π2cos⁡x2x−πdx.

Answer:

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77. Evaluate $\int_{0}^{π} \frac{x \sin (2 x) \sin \frac{π}{2} \cos x}{2 x - π} d x$ .