Evaluate: `int_0^(pi/4)(tan^(-1)((2cos^2theta)/(2-sin2theta)))sec^2thetad``theta`.

Explanation

Given integral is `int_(0)^(pi//4)"tan"^(-1)((2cos^(2)theta)/(2-sin2theta))sec^(2) theta d theta`
`=int_(0)^(pi//4)"tan"^(-1)(1/(sec^(2)theta-tan theta))sec^(2) theta d theta`
`=int_(0)^(pi//4)"tan"^(-1)(1/(1+tan^(2)theta-tan theta))sec^(2) theta d theta`
Put `tan theta=t`. Then `sec^(2)theta d theta=dt`
The given integral reduces to
`int_(0)^(1)"tan"^(-1)(1/(1+t^(2)-1))dt=int_(0)^(1)"tan"^(-1)((t-(t-1))/(1+t(t-1)))dt`
`=int_(0)^(1)"tan"^(-1)t dt -int_(0)^(1)tan^(-1)(t-1)dt`
`=int_(0)^(1) tan^(-1)tdt-int_(0)^(1)tan^(-1)((1-t)-1)dt`
`=2int_(0)^(1)1tan^(-1)t dt`
`=2[t tan^(-1)t]_(0)^(1)-2int_(0)^(1)t/(1+t^(2))dt`
(Integrating by parts)
`(pi)/2-[In(1+t^(2))]_(0)^(1)=(pi)/2-In2`.