Evaluate: `int(cosx)/(sqrt(sin^2x-2sinx-3))\ dx`

Explanation

I= `int(cosx)/(sqrt(sin^2x-2sinx-3))\ dx`

I= `int(cosx)/(sqrt((sinx-1)^2sinx-2^2))\ dx`

Let `sinx−1=t`

`⇒cosx.dx=dt`

Substituting we get-

`I=int (dt)/(sqrt(t^2-2^2))`

`=log∣t+sqrt(t^2−2^2)​∣​+C`

`=log∣​(sinx−1)+sqrt((sinx−1)^2−4
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