Question

If $f$ is an even function, then the value of $\underset{0}{\overset{\pi /2}{\int }}f\left(\cos 2x\right)\cos xdx$ can be equal to

• A. $2\underset{0}{\overset{\pi /4}{\int }}f\left(\sin 2t\right)\cos tdt$
• B. $\sqrt{2}\underset{0}{\overset{\pi /4}{\int }}f\left(\sin 2t\right)\cos tdt$
• C. $2\underset{0}{\overset{\pi /4}{\int }}f\left(\sin 2t\right)\sin tdt$
• D. $\sqrt{2}\underset{0}{\overset{\pi /4}{\int }}f\left(\sin 2t\right)\sin tdt$

Answer 1

The correct option is B $\sqrt{2}\underset{0}{\overset{\pi /4}{\int }}f\left(\sin 2t\right)\cos tdt$

Let $I=\underset{0}{\overset{\pi /2}{\int }}f\left(\cos 2x\right)\cos xdx\cdots \left(i\right)$
Using property
$\underset{a}{\overset{b}{\int }}f\left(x\right)dx=\underset{a}{\overset{b}{\int }}f(a+b-x)dx$
$\Rightarrow I=\underset{0}{\overset{\pi /2}{\int }}f(-\cos 2x)\sin xdxI=\underset{0}{\overset{\pi /2}{\int }}f\left(\cos 2x\right)\sin xdx\cdots \left(ii\right)$
As, $f$ is even (given)
Adding $\left(i\right)$ and $\left(ii\right):$
$2I=\sqrt{2}\underset{0}{\overset{\pi /2}{\int }}f\left(\cos 2x\right)\cos (x-\frac{\pi }{4})dx$
Substitute, $-x+\frac{\pi }{4}=t,-dx=dt$
$\Rightarrow 2I=-\sqrt{2}\underset{\pi /4}{\overset{-\pi /4}{\int }}f\left(\cos (\frac{\pi }{2}-2t)\right)\cos tdt\Rightarrow I=\sqrt{2}\underset{0}{\overset{\pi /4}{\int }}f\left(\sin 2t\right)\cos tdt[\because f\text{is even}]$