Indefinite Integration 1 Question 77
78. Show that,
$\int _0^{\pi / 2} f(\sin 2 x) \sin x d x=\sqrt{2} \int _0^{\pi / 4} f(\cos 2 x) \cos x d x$.
$(1990,4 M)$
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Solution:
- Let $I=\int _0^{\pi / 2} f(\sin 2 x) \sin x d x$
Then, $I=\int _0^{\pi / 2} f \sin 2 \frac{\pi}{2}-x \quad \sin \frac{\pi}{2}-x d x$
$$ =\int _0^{\pi / 2} f[\sin 2 x] \cdot \cos x d x $$
On adding Eqs. (i) and (ii), we get
$$ \begin{aligned} & 2 I=\int _0^{\pi / 2} f(\sin 2 x)(\sin x+\cos x) d x \\ &=2 \int _0^{\pi / 4} f(\sin 2 x)(\sin x+\cos x) d x \\ &=2 \sqrt{2} \int _0^{\pi / 4} f(\sin 2 x) \sin x+\frac{\pi}{4} d x \\ &=2 \sqrt{2} \int _0^{\pi / 4} f \sin 2 \frac{\pi}{4}-x \text { sin } \frac{\pi}{4}-x+\frac{\pi}{4} d x \\ &=2 \sqrt{2} \int _0^{\pi / 4} f(\cos 2 x) \cos x d x \\ & \therefore \quad I=\sqrt{2} \int _0^{\pi / 4} f(\cos 2 x) \cos x d x \\ & \text { Hence, } \int _0^{\pi / 2} f(\sin 2 x) \cdot \sin x d x=\sqrt{2} \int _0^{\pi / 4} f(\cos 2 x) \cos x d x \end{aligned} $$
