The value of `int_(-pi//2)^(pi//2)(sin^(2)x)/(1+2^(x))dx` is

Options

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

Answer

D

Explanation

Key idea Use property `=int_(a)^(b)f(x)dx=int_(a)^(b)f(a+b-x)dx`
Let `=int_(-pi//2)^(pi//2)(sin^(2)x)/(1+2^(x))dx` `rArrI=int_(-pi//2)^(pi//2)sin^(2)(-(pi)/(2)+(pi)/(2)-x)/(1+2^((pi)/(2)+(pi)/(2)-x))dx`
`[:'int_(a)^(b)f(x)dx=int_(a)^(b)f(a+b-x)dx]`
`rArrI=int_(-pi//2)^(pi//2)(sin^(2)x)/(1+2^(-x))dx`
`rArrI=int_(-pi//2)^(pi//2)(2^(x)sin^(2)x)/(2^(x)+1)dx`
`rArr2I=int_(-pi//2)^(pi//2)sin^(2)x((2^(x)+1)/(2^(x)+1))dx`
`rArr2I=int_(-pi//2)^(pi//2)sin^(2)x dx`
`rArr2I= 2 int_(0)^(pi//2)sin^(2)xdx[:'sin^(2)x " is an even function"]`
`rArrI=int_(0)^(pi//2)sin^(2) x dx`
`rArrI= int_(0)^(pi//2)cos^(2)x dx [ :'int_(0)^(a)f (x) dx=int_(0)^(a)f(a-x)dx]`
`rArr2I=int_(0)^(pi//2)dx`
`rArr2I=[x]_(0)^(pi//2)rArrI=(pi)/(4)`