Question

Show that, $\int\limits_{0}^{\pi / 2} f(\sin 2 x) \sin x d x=\sqrt{2} \int\limits_{0}^{\pi / 4} f(\cos 2 x) \cos x d x$

Answer 1

Let $ I=\int\limits_{0}^{\pi / 2} f(\sin 2 x) \sin x d x ....$(i)
Then, $I=\int\limits_{0}^{\pi / 2} f\left[\sin 2\left(\frac{\pi}{2}-x\right)\right] \sin \left(\frac{\pi}{2}-x\right) d x$
$=\int\limits_{0}^{\pi / 2} f[\sin 2 x] \cdot \cos x d x ....$(ii)
On adding Eqs. (i) and (ii), we get
$2 I =\int\limits_{0}^{\pi / 2} f(\sin 2 x)(\sin x+\cos x) d x $
$=2 \int\limits_{0}^{\pi / 4} f(\sin 2 x)(\sin x+\cos x) d x $
$=2 \sqrt{2} \int\limits_{0}^{\pi / 4} f(\sin 2 x) \sin \left(x+\frac{\pi}{4}\right) d x $
$=2 \sqrt{2} \int\limits_{0}^{\pi / 4} f\left(\sin 2\left(\frac{\pi}{4}-x\right)\right) \sin \left(\frac{\pi}{4}-x+\frac{\pi}{4}\right) d x $
$=2 \sqrt{2} \int\limits_{0}^{\pi / 4} f(\cos 2 x) \cos x d x$
$\therefore I=\sqrt{2} \int\limits_{0}^{\pi / 4} f(\cos 2 x) \cos x d x$
Hence, $\int\limits_{0}^{\pi / 2} f(\sin 2 x) \cdot \sin x d x=\sqrt{2} \int\limits_{0}^{\pi / 4} f(\cos 2 x) \cos x d x$