Question

Show that, $\underset{0}{\overset{\pi /2}{\int }}f(\sin 2x)\sin xdx=\sqrt{2}\underset{0}{\overset{\pi /4}{\int }}f(\cos 2x)\cos xdx$

Answer 1

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