Question

If $I_{1}=\int_{0}^{\frac{\pi}{2}} f (sin \,2x)sin \,xdx$ and $I_2 = \int^{\pi/4}_{0}\,f(cos2x) cosx \,dx\,then\, I_1/I_2$ =

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

Answer 1

Answer: (B)

$\sqrt{2}$

Answer 2

The correct option is(B): √2.

Given $I_1 = \int\limits_0^{\pi/2} f(sin\,2x) sin\,x dx$
$\Rightarrow I_1 = \int\limits_0^{\pi/2} f(sin\,2x) cos\,x dx$
$(\because \int\limits_0^a f(x) dx = \int\limits_0^a f( a-x) dx ]$
$\Rightarrow 2I_1 = \int\limits_0^{\pi/2} f(sin\,2x)(sin\,x) + cos\,x )dx$
$= \sqrt{2} \int\limits_0^{\pi/2} f(sin\,2x) cos(x - \frac{\pi}{4})dx$
Put $x - \frac{\pi}{4} = t$
$\Rightarrow dx = dt$
$\therefore 2I_1 = \sqrt{2} \int\limits_{-\pi/4}^{\pi/4} f(sin (\frac{\pi}{2} + 2t)) cos\,t \,dt$
$\therefore 2I_1 = 2\sqrt{2}\int\limits_0^{\pi/4} f(cos\,2t) cos\,t\,dt$
$\Rightarrow I_1 = \sqrt{2}\int\limits_0^{\pi/4} f(cos\,2x) \,cos\,x\,dx$
$\Rightarrow I_1 = \sqrt{2}$