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The value of int-pi^pi cos^2x/[1+a^x].dx...

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

Text Solution

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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,
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