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

The Correct Option is B

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