The value of `int_-pi^pi cos^2x/[1+a^x].dx`,a>0 is

Answer

`pi//2`

Explanation

Let `I=int_(-pi)^(pi)(cos^(2)x)/(1+a^(x))dx, agt0` ………….. 1
`:.I=int_(-pi)^(pi)(cos^(2)x)/(1+a^(-x))dx` (Replacing `x` by `-x`)
or `I=int_(-pi)^(pi)(a^(x)cos^(2)x)/(1+a^(x))dx`…………….2
Adding 1 and 2 we get
`2I=4int_(0)^(pi//2)cos^(2)x dx`
`=2int_(0)^(pi//2)(1+cos2x)dx`
`=2[x+(sin2x)/2]_(0)^(pi//2)`
`-pi`
or `I=(pi)/2`