Evaluate the following integral.

{eq}\displaystyle \int \dfrac { \cot x \ dx }{ \csc x + 16 \sin x} {/eq}

Given:

{eq}\displaystyle \int\ \frac{\cot\ x}{\csc\ x + 16\sin\ x}\ dx {/eq}

Expand the cotangent term. Recall:

{eq}\displaystyle \cot\ x = \frac{\cos\ x}{\sin\ x} {/eq}

{eq}\displaystyle \int\ \frac{\cot\ x}{\csc\ x + 16\sin\ x}\ dx = \int\ \frac{\cos\ x}{\sin\ x(\csc\ x + 16\sin\ x)}\ dx {/eq}

{eq}\displaystyle \int\ \frac{\cot\ x}{\csc\ x + 16\sin\ x}\ dx = \int\ \frac{\cos\ x}{\sin\ x\csc\ x + 16\sin^2\ x}\ dx {/eq}

Recall that:

{eq}\displaystyle \sin\ x\csc\ x = 1 {/eq}

Therefore,

{eq}\displaystyle \int\ \frac{\cot\ x}{\csc\ x + 16\sin\ x}\ dx = \int\ \frac{\cos\ x}{1 + 16\sin^2\ x}\ dx {/eq}

Let:

{eq}\displaystyle u = 4\sin\ x {/eq}

{eq}\displaystyle du = 4\cos\ x\ dx {/eq}

Or:

{eq}\displaystyle \frac{1}{4}\ du = \cos\ x\ dx {/eq}

In terms of u, we can write this integral as:

{eq}\displaystyle \int\ \frac{\cot\ x}{\csc\ x + 16\sin\ x}\ dx = \frac{1}{4}\int\ \frac{1}{1+ u^2}\ du {/eq}

This is a standard integral:

{eq}\displaystyle \int\ \frac{1}{1+ u^2}\ du = \tan^{-1}\ u + C {/eq}

Therefore,

{eq}\displaystyle \int\ \frac{\cot\ x}{\csc\ x + 16\sin\ x}\ dx = \frac{1}{4}\tan^{-1}\ u + C {/eq}

Return the definition of u to get:

{eq}\displaystyle \boxed{\int\ \frac{\cot\ x}{\csc\ x + 16\sin\ x}\ dx = \frac{1}{4}\tan^{-1}\ (4\sin\ x) + C} {/eq}