`int_0^(pi/2) (xsinxcosx)/(cos^4x+sin^4x) dx=`

Options

A. `(pi^(2))/2`
B. `(pi^(2))/4`
C. `(pi^(2))/8`
D. `(pi^(2))/16`

Answer

D

Explanation

Let
`I=int_(0)^(pi//2)(x sin x cos x)/(cos^(4)x +sin^(4)x)dx`…………….. 1
`implies I=int_(0)^(pi//2) ((pi//2-x)sinx cos x)/(sin^(4)x+cos^(4)x) dx`…………2
Adding 1 and 2 we get
`I=(pi)/4int_(0)^(pi//2) (sinxcosx)/(sin^(4)x+cos^(4)x)dx`
`=(pi)/4 int_(0)^(pi//2)(sec^(2)x tan x)/(tan^(4)x+1)dx` (Dividing Nr. And Dr. by `cos^(4)x`)
`=(pi)/(2xx4)int_(0)^(pi//2)((tan^(2)x)'dx)/(1+(tan^(2)x)^(2))`
`=(pi)/8[tan^(-1)(tan^(2)x)]_(0)^(pi//2)`
`=(pi)/8[(pi)/2-0]`
`=(pi^(2))/16`