Question

If ${\int }_0^{\pi /2}f\left(\sin 2x\right)\sin xdx=A\frac{\sqrt{2}}{9}{\int }_0^{\pi /4}f\left(\cos 2x\right)\cos xdx$ then the value of $A$ is

Answer 1

${\int }_0^{\pi /2}f\left(\sin 2x\right)\sin xdx={\int }_0^{\pi /2}f\left(\sin \right(\pi -2x\left)\right)\sin (\frac{\pi }{2}-x)dx{\int }_0^{\pi /2}f\left(\sin 2x\right)\cos xdx$ ...(1)

${\int }_0^{\pi /4}f\left(\cos 2x\right)\cos xdx={\int }_0^{\pi /4}f\left(\cos (\frac{\pi }{2}-2x)\right)\cos (\frac{\pi }{4}-x)dx$

$={\int }_0^{\pi /4}f\left(\sin 2x\right)(\frac{1}{\sqrt{2}}\cos x+\frac{1}{\sqrt{2}}\sin x)dx$ ...(2)

${\int }_0^{\pi /2}f\left(\sin 2x\right)(\sin x+\cos x)dx=2{\int }_0^{\pi /4}f\left(\sin 2x\right)(\sin x+\cos x)dx$ ...(3)
Therefore
$\sqrt{2}{\int }_0^{\pi /4}f\left(\cos 2x\right)\cos xdx={\int }_0^{\pi /4}f\left(\sin 2x\right)(\cos x+\sin x)dx$ (From(2))

$=\frac{1}{2}{\int }_0^{\pi /2}f\left(\sin 2x\right)(\sin x+\cos x)dx$ (From(3))

$=\frac{1}{2}\cdot 2{\int }_0^{\pi /2}f\left(\sin 2x\right)\sin xdx$ (From (1))

$={\int }_0^{\pi /2}f\left(\sin 2x\right)\sin xdx$
Hence $A=9$