Question

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

Answer 1

$I=\underset{0}{\overset{\pi /2}{\int }}f\left(\sin 2x\right)\sin xdx$ ..........(1)

$I=\underset{0}{\overset{\pi /2}{\int }}f\left[\sin 2\right(\frac{\pi }{2}-x\left)\right]\sin (\frac{\pi }{2}-x)dx$ $\left(\underset{a}{\overset{b}{\int }}f\right(x)dx=\underset{a}{\overset{b}{\int }}f(a+b-x\left)dx\right)$

$I=\underset{0}{\overset{\pi /2}{\int }}f\left[\sin \right(\pi -2x\left)\right]\cos xdx$

$I=\underset{0}{\overset{\pi /2}{\int }}f\left(\sin 2x\right)\cos xdx$ ........(2)

On adding equation 1 and 2, we get,

$2I=\underset{0}{\overset{\pi /2}{\int }}f\left(\sin 2x\right).(\sin x+\cos x)dx$

$2I=2\underset{0}{\overset{\pi /4}{\int }}f\left(\sin 2x\right).(\sin x+\cos x)dx$ $\left(\underset{0}{\overset{2a}{\int }}f\right(x)dx=\underset{0}{\overset{a}{\int }}2f(x)dx,iff(2a-x)=f(x\left)\right)$

$2I=2\sqrt{2}\underset{0}{\overset{\pi /4}{\int }}f\left(\sin 2x\right)(\frac{1}{\sqrt{2}}\sin x+\frac{1}{\sqrt{2}}\cos x)dx$

$2I=2\sqrt{2}\underset{0}{\overset{\pi /4}{\int }}f\left(\sin 2x\right)\sin (x+\frac{\pi }{4})dx$

$2I=2\sqrt{2}\underset{0}{\overset{\pi /4}{\int }}f\left[\sin 2\right(\frac{\pi }{4}-x\left)\right]\sin (\frac{\pi }{4}-x+\frac{\pi }{4})dx$

$2I=2\sqrt{2}\underset{0}{\overset{\pi /4}{\int }}f\left[\sin \right(\frac{\pi }{2}-2x\left)\right]\sin (\frac{\pi }{2}-x)dx$

$2I=2\sqrt{2}\underset{0}{\overset{\pi /4}{\int }}f\left(\cos 2x\right)\cos xdx$

$I=\sqrt{2}\underset{0}{\overset{\pi /4}{\int }}f\left(\cos 2x\right)\cos xdx$

Hence, proved.