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

Explanation

Given


`I=int_{0}^{x} frac{d x}{(6-cos x)}`

`
cos x=frac{1+tan ^{2}(frac{z}{2})}{1+tan ^{2}(frac{z}{2})}, `

we obtain

`

I =int_{0}^{pi} frac{1}{6-frac{1-tan ^{2}(frac{x}{2})}{1+tan ^{2}(frac{x}{2})}} d x `

`
=int_{0}^{pi} frac{sec ^{2}(frac{x}{2})}{5+7 tan ^{2}(frac{x}{2})} d x`



Take `tan (frac{x}{2})=t`

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