Integral of csc^4(4x) dx

1. Rewrite the integrand:

   $$\csc^{4}{\left(4 x \right)} = \left(\cot^{2}{\left(4 x \right)} + 1\right) \csc^{2}{\left(4 x \right)}$$

2. Let $u = \cot{\left(4 x \right)}$.

   Then let $du = \left(- 4 \cot^{2}{\left(4 x \right)} - 4\right) dx$ and substitute $du$:

   $$\int \left(- \frac{u^{2}}{4} - \frac{1}{4}\right)\, du$$

   1. Integrate term-by-term:

      1. The integral of a constant times a function is the constant times the integral of the function:

         $$\int \left(- \frac{u^{2}}{4}\right)\, du = - \frac{\int u^{2}\, du}{4}$$

         1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$:

            $$\int u^{2}\, du = \frac{u^{3}}{3}$$

         So, the result is: $- \frac{u^{3}}{12}$

      1. The integral of a constant is the constant times the variable of integration:

         $$\int \left(- \frac{1}{4}\right)\, du = - \frac{u}{4}$$

      The result is: $- \frac{u^{3}}{12} - \frac{u}{4}$

   Now substitute $u$ back in:

   $$- \frac{\cot^{3}{\left(4 x \right)}}{12} - \frac{\cot{\left(4 x \right)}}{4}$$

3. Now simplify:

   $$- \frac{\left(3 + \frac{1}{\tan^{2}{\left(4 x \right)}}\right) \cot{\left(4 x \right)}}{12}$$

4. Add the constant of integration:

   $$- \frac{\left(3 + \frac{1}{\tan^{2}{\left(4 x \right)}}\right) \cot{\left(4 x \right)}}{12}+ \mathrm{constant}$$

The answer is: