# Given the eq f(x)+f(-x)=pi, show how calculate definite integral (0 t0 3pi) f(cos x) dx?

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### 1 Answer

We know only one fact about `f(x)` , which is that `f(x)+f(-x)=pi` . Now when we are calculating the integral, we see that somehow we need to use this property, which suggests that we divide up the domain in two.

`int_0^{3pi} f(cos x)dx`

`=int_0^{{3pi}/2} f(cosx)dx+int_{{3pi}/2}^{3pi}f(cos x)dx` but `cos x` is odd around the point `x={3pi}/2` which means that the second integral can change its limits and argument to get

`=int_0^{{3pi}/2} f(cos x)dx + int_0^{{3pi}/2}f(-cos x)dx`

`=int_0^{{3pi}/2} (f(cos x)+f(-cos x))dx` now use the property of `f(x)`

`=int_0^{{3pi}/2} pi dx` now integrate

`=pi (x)_0^{{3pi}/2}`

`=3/2 pi^2`

**The integral evaluates to `3/2 pi^2` . **