Let `f` beintegrable over `[0,a]` for any real value of `a`.
If `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

Options

A. `I_(1)=-2I_(2)`
B. `I_(1)=I_(2)`
C. `2I_(1)=I_(2)`
D. `I_(1)=-I_(2)`

Answer

B

Explanation

`I_(1)-I_(2)=int_(0)^(pi//2) (cos theta -sin 2 theta)f(sin theta+cos^(2) theta) d theta`
Put `t=sin theta +cos^(2) theta` and `dt=(cos theta-sin 2 theta)d theta`
or `I_(1)-I_(2)=int_(1)^(1)f(t)dt=0`