1.

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

Answer» Here, `I =f(x) = int_(-pi)^(pi)cos^2x/(1+a^x)dx->(1)`
`I = f(-x) = int_(-pi)^(pi) cos^2(-x)/(1+a^-x)dx`
`I= int_(-pi)^(pi) cos^2 x/(1+1/a^x)dx`

`I= int_(-pi)^(pi) (a^x cos^2x)/(1+a^x)dx->(2)`

Adding (1) and (2),
`2I = int_(-pi)^(pi) (cos^2 x)(1+a^x)/(1+a^x)`
`=> 2I = int_(-pi)^(pi) (cos^2 x)->(3)`
Now, as `cos^2 x` is an even function,
`:. int_(-pi)^(pi) (cos^2 x) =2 int_(0)^(pi) (cos^2 x)`
So, (3) becomes,
`:. 2I = 2 int_(0)^(pi) (cos^2 x)`
Now, using `cos2x = 2cos^2x -1` in our expression,
`=> I = int_(0)^(pi) 1/2 (1+cos2 x)`
`=> I =1/2(x+ 2sin(2x)/2)_(0)^(pi)`
`=>1/2(pi-0) = pi/2`
`:. I = pi/2`


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