Q. Let $f$ be integrable over $\lbrack 0,a\rbrack$ for any real $a.$ If we define $I_{1} = \int_{0}^{\pi/2}{\cos{\theta f(\sin{\theta + \cos^{2}{\theta)\ d\theta}}}}$ and $I_{2} = \int_{0}^{\pi/2}{\sin{2\theta f(\sin{\theta + \cos^{2}{\theta)\ d\theta,}}}}$ then
A) $I_{1} = I_{2}$
B) $I_{1} = - I_{2}$
C) $I_{1} = 2I_{2}$
D) $I_{1} = {- 2I}_{2}$
Explanation
(a)
$I_{1} - I_{2} = \int_{0}^{\pi/2}{(\cos{\theta - \sin{2\theta)f(\sin{\theta + \cos^{2}{\theta)d\theta}}}}}$
Put $\sin{\theta + \cos^{2}{\theta = t}}$
$\Rightarrow (\cos{\theta - \sin{2\theta)d\theta = dt}}$
Then, $I_{1} - I_{2} = \int_{1}^{1}{f(t)dt = 0}$
$\therefore\ I_{1} = I_{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.