Evaluate the following integral: `int_0^pi(dx)/(6-cosx)`

Explanation

Given:`int_0^pi(dx)/(6-cosx)`

`I=int_0^pi(dx)/(6-cosx)`

`I=int_0^pi(dx)/(6-(1-tan^2(x/2))/(1+tan^2(X/2)))dx`

`I=int_0^pi(sec^2(x/2))/(6+6tan^2(x/2)-1+tan^2(x/2))`

`I=int_0^pi(sec^2(x/2))/(5+7tan^2(x/2))dx`

`I=1/7int_0^pi(sec^2(x/2))/(5/7+tan^2(x/2))dx`

Let `tan(x/2)=t implies 1/2sec^2(x/2)dx=dt implies sec^2(x/2)dx=2dt`

`I=2/7int_0^(oo) 1/(((sqrt(5/7))^2+t^2))dx`

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