Question

If $I_1={\int }_0^{\frac{\pi }{2}}f\left(\sin 2x\right)\sin xdx$ and $I_2={\int }_0^{\frac{\pi }{2}}f\left(\cos 2x\right)\cos xdx$, then find $\frac{I_1}{I_2}$.

Answer 1

$I_1={\int }_0^{\frac{\pi }{2}}f\left(\sin 2x\right)\sin xdx$

$={\int }_0^{\frac{\pi }{4}}f\left(\sin 2x\right)\sin xdx+{\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}f\left(\sin 2x\right)\sin xdx$

Setting $x=\frac{\pi }{4}-t$ in the first integral and $x=\frac{\pi }{4}+t$ in the second integral,

$I_1={\int }_0^{\frac{\pi }{4}}f\left(\sin (\frac{2\pi }{4}-2t)\right)\sin (\frac{\pi }{4}-t)dt+{\int }_0^{\frac{\pi }{4}}f\left(\cos 2t\right)\sin (\frac{\pi }{4}+t)dt$

$={\int }_0^{\frac{\pi }{4}}f\left(\cos 2t\right)[\sin (\frac{\pi }{4}-t)+\sin (\frac{\pi }{4}+t)]dt$

$={\int }_0^{\frac{\pi }{4}}f\left(\cos 2t\right)2\sin \left(\frac{\pi }{4}\right)\cos tdt$

$=\sqrt{2}{\int }_0^{\frac{\pi }{4}}f\left(\cos 2t\right)\cos tdt$

We have, ${\int }_0^{nT}f\left(x\right)dx=n{\int }_0^{T}f\left(x\right)dx$

$I_1=\frac{\sqrt{2}}{2}{\int }_0^{\frac{\pi }{2}}f\left(\cos 2t\right)\cos tdt$

$=\frac{1}{\sqrt{2}}{I}_2$

$\Rightarrow \frac{I_1}{I_2}=\frac{1}{\sqrt{2}}$