Evaluate the following integral: `int_0^(pi//2)s in2xtan^(-1)(sinx)dx`

Explanation

Let,I=`int_0^(pi//2)sin2xtan^(-1)(sinx)dx`


I=`int_0^(pi//2) 2sinxcosxtan^(-1)(sinx)dx`


Let sinx=t⇒cosxdx=dt

when x=0,t=0

when x=`π/2​,t=1`

=`int_0^1​.2tcosxtan^(−1)(t).dt/cosx`
​
...