The value of the integral `int_(-pi//2)^(pi//2) (x^(2) + log" (pi-x)/(pi+x)) cos x dx `

Options

A. 0
B. `(pi^(2))/(2)-4`
C. `(pi^(2))/(2)-4`
D. `(pi^(2))/(2)`

Answer

B

Explanation

Let`I=overset(pi//2)underset(-pi//2)int{x^(2)+log_(e )((pi+x)/(pi-x))}cos x dx`.Then,
`I=overset(pi//2)underset(-pi//2)int{x^(2)+log_(e )((pi+x)/(pi-x))}cos x dx`.Then,
`I=overset(pi//2)underset(-pi//2)int x^(2)cos c dx+overset(pi//2)underset(-pi//2)int log((pi+x)/(pi-x))cos x dx`
`rArr I=2overset(pi//2)underset(0)int underset(I)(x^(2))cos xdx+0" "[:' log((pi+x)/(pi-x))cos x
"is an odd function"]`
`I=2[x^(2)sinx+2x cos x-2 sinx]_(0)^(pi//2)`
`rArr I=2((pi^(2))/(4)-2)=(pi^(2))/(2)-4`