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)

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$
$\Rightarrow dx = dt$
$\therefore 2I_1 = \sqrt{2} \int\limits_{-\pi/4}^{\pi/4} f(sin (\frac{\pi}{2} + 2t)) cos\,t \,dt$
$\therefore 2I_1 = 2\sqrt{2}\int\limits_0^{\pi/4} f(cos\,2t) cos\,t\,dt$
$\Rightarrow I_1 = \sqrt{2}\int\limits_0^{\pi/4} f(cos\,2x) \,cos\,x\,dx$
$\Rightarrow I_1 = \sqrt{2}I_2$