Question

If $f\left(x\right)$ is an even function, then prove that ${\int }_0^{\pi /2}f\left(\cos 2x\right)\cos xdx=\sqrt{2}{\int }_0^{\pi /4}f\left(\sin 2x\right)\cos xdx$.

Answer 1

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

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

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

$={\int }_0^{\frac{\pi }{2}}f(-\cos 2x)\sin xdx$

$I=-{\int }_0^{\frac{\pi }{2}}f\left(\cos 2x\right)\sin xdx$ $[\because f(x)$ is even function]

equation (1)+(2)

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

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

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

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

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

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

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

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

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

$I=\sqrt{2}{\int }_0^{\frac{\pi }{4}}+(\cos 2(\frac{\pi }{4}-x))\left|\sin (\frac{\pi }{4}-(\frac{\pi }{4}-x))\right|$

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

$=\sqrt{2}{\int }_0^{\frac{\pi }{2}}f\left(\sin x\right)\sin xdx$