Evaluate: `int1/(sinxcos^2x)dx`

Explanation

Let `u=tan ^{-1}(frac{cos x}{1+sin x}) and v=sec ^{-1} x.`

Then we want to find `frac{d u}{d v}.`

Differentiate u and v w.r.t. `x_{t}` we get


`frac{d u}{d x}=frac{d}{d x}[tan ^{-1}(frac{cos }{1+sin x})]`


`frac{cos x}{1+sin x}=frac{sin (frac{pi}{2}-x)}{1+cos (frac{pi}{2}-x)} `
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