Question

If $I_1={\int }_0^{\frac{\pi }{2}}f\left(\sin 2x\right)\sin xdx$
and $I_2={\int }_0^{\frac{\pi }{4}}f\left(\cos 2x\right)\cos xdx,$ then $\frac{I_1}{I_2}$ is equal to

• A. $1$
• B. $\frac{1}{\sqrt{2}}$
• C. $\sqrt{2}$
• D. $2$

Answer 1

Answer: (C)

$\sqrt{2}$

$I_1={\int }_0^{\frac{\pi }{2}}f\left(\sin 2x\right)\sin xdx$ ...(1)

Using ${\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f(a-x)dx,$ we get
$I_1={\int }_0^{\frac{\pi }{2}}f\left(\sin \right(\pi -2x\left)\right)\sin (\frac{\pi }{2}-x)dx$
$I_1={\int }_0^{\frac{\pi }{2}}f\left(\sin 2x\right)\cos xdx$ ...(2)

Adding $\left(1\right)$ and $\left(2\right)$ we get
$2I_1={\int }_0^{\frac{\pi }{2}}f\left(\sin 2x\right)(\sin x+\cos x)dx$
$2I_1=\sqrt{2}{\int }_0^{\frac{\pi }{2}}f\left(\sin 2x\right)\cos (x-\frac{\pi }{4})dx$

Put $x-\frac{\pi }{4}=\theta \Rightarrow dx=d\theta$, so that
$2I_1=\sqrt{2}{\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}f\left(\sin (\frac{\pi }{2}+2\theta )\right)\cos \theta d\theta$
$2I_1=\sqrt{2}{\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}f\left(\cos 2\theta \right)\cos \theta d\theta =2\sqrt{2}{\int }_0^{\frac{\pi }{4}}f\left(\cos 2\theta \right)\cos \theta d\theta$

As integrand is an even function

$I_1=\sqrt{2}{I}_1$

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