Evaluate the following integral: `int_0^(pi//2)2s in x\ cos xtan^(-1)(sinx)dx`

Explanation

Let I=`int_0^(pi//2)2s in x\ cos xtan^(-1)(sinx)dx`

Put sinx=z⇒cosxdx=dz

The limits are, when
`x=0,z=sin0=0;x=π/2​,z=sin(π/2)​=1`

I=`2int_0^(1)ztan^(−1)(z)dz`

=`2[tan^(−1)z⋅(z^2/2)​]_0^1​−2int_0^1 ​(z^2​dz)/(1+z^2)`


=`pi/4-int_0^1(1+z^2-1)/(1+z^2)dz`

=`pi/4-1+pi/4`

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