SATHEE: Indefinite Integration 1 Question 77

Indefinite Integration 1 Question 77

78. Show that,

$\int_{0}^{π / 2} f (\sin 2 x) \sin x d x = \sqrt{2} \int_{0}^{π / 4} f (\cos 2 x) \cos x d x$ .

$(1990, 4 M)$

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Solution:

Let $I = \int_{0}^{π / 2} f (\sin 2 x) \sin x d x$

Then, $I = \int_{0}^{π / 2} f \sin 2 \frac{π}{2} - x \sin \frac{π}{2} - x d x$

$= \int_{0}^{π / 2} f [\sin 2 x] \cdot \cos x d x$

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

$\begin{aligned} 2 I = \int_{0}^{π / 2} f (\sin 2 x) (\sin x + \cos x) d x \\ = 2 \int_{0}^{π / 4} f (\sin 2 x) (\sin x + \cos x) d x \\ = 2 \sqrt{2} \int_{0}^{π / 4} f (\sin 2 x) \sin x + \frac{π}{4} d x \\ = 2 \sqrt{2} \int_{0}^{π / 4} f \sin 2 \frac{π}{4} - x sin \frac{π}{4} - x + \frac{π}{4} d x \\ = 2 \sqrt{2} \int_{0}^{π / 4} f (\cos 2 x) \cos x d x \\ ∴ I = \sqrt{2} \int_{0}^{π / 4} f (\cos 2 x) \cos x d x \\ Hence, \int_{0}^{π / 2} f (\sin 2 x) \cdot \sin x d x = \sqrt{2} \int_{0}^{π / 4} f (\cos 2 x) \cos x d x \end{aligned}$

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Indefinite Integration 1 Question 77

78. Show that,

Solution:

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