Evaluate: `int_(-pi/2)^(pi/2)sin^2xcos^2x(sinx+cosx)dx`

Answer

`4//15`

Explanation

Let `=int_(-pi//2)^(pi//2)sin^(2)x cos^(2)x(sinx+cosx)dx`
`=int_(-pi//2)^(pi//2) sin^(3)x cos^(2)x dx+int_(-pi//2)^(pi//2) sin^(2) x cos^(3)x dx`………….1
Since `sin^(3)x cos^(2)x` is an odd functional and `sin^(2)x cos^(3)x` is an even function. Therefore `int_(-pi//2)^(pi//2) sin^(3)x cos^(2)x dx=0`
and `int_(-pi//2)^(pi//2) sin^(2) x cos^(3)x dx =int_(0)^(pi//2) sin^(2)x cos^(3)x dx`.
`:. I=2int_(0)^(pi//2) sin^(2)x cos^(3)x dx`
`=2int_(0)^(1)t^(2)(1-t^(2))dt`
`=int_(0)^(1)(t^(2)-t^(4))dt=2[1/3-1/5]=4/15`