The value of `int_(-pi/2)^(pi/2) (x^2cosx)/(1+e^x) dx` is equal to

Options

A. `(pi^(2))/4-2`
B. `(pi^(2))/4+2`
C. `pi^(2)-e^((pi)/2)`
D. `pi^(2)+e^(pi)/2`

Answer

A

Explanation

`I=int_(-(pi)/2)^((pi)/2)(x^(2)cosx)/(1+e^(x))dx`………………1
`=int_(-(pi)/2)^((pi)/2)(x^(2)cosx)/(1+1/(e^(x))`(Replacing x by `-x`)
`:. I=int_(-(pi)/2)^((pi)/2)(x^(2)cosx.e^(x))/(1+e^(x))dx`……………..2
Adding 1 and 2 we get
`2I=int_(-(pi)/2)^((pi)/2)x^(2)cosx dx`
`:. I=int_(0)^((pi)/2)x^(2)cosxdx`
`=x^(2). sin x|_(0)^(pi//2)-int_(0)^((pi)/2)2x sin xdx`
`=(pi^(2))/4-2[(-xcosx)_(0)^(pi//2)-int_(0)^(pi//2)( -cosx)dx]`
`=(pi^(2))/4-2[1]=(pi^(2))/4-2`