Integral of sin^4(2x)cos(2x) dx

1. There are multiple ways to do this integral.

   Method #1

   1. Let $u = \sin{\left(2 x \right)}$.

      Then let $du = 2 \cos{\left(2 x \right)} dx$ and substitute $\frac{du}{2}$:

      $$\int \frac{u^{4}}{4}\, du$$

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

         $$\int \frac{u^{4}}{2}\, du = \frac{\int u^{4}\, du}{2}$$

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

            $$\int u^{4}\, du = \frac{u^{5}}{5}$$

         So, the result is: $\frac{u^{5}}{10}$

      Now substitute $u$ back in:

      $$\frac{\sin^{5}{\left(2 x \right)}}{10}$$

   Method #2

   1. Let $u = 2 x$.

      Then let $du = 2 dx$ and substitute $\frac{du}{2}$:

      $$\int \frac{\sin^{4}{\left(u \right)} \cos{\left(u \right)}}{4}\, du$$

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

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

         1. Let $u = \sin{\left(u \right)}$.

            Then let $du = \cos{\left(u \right)} du$ and substitute $du$:

            $$\int u^{4}\, du$$

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

               $$\int u^{4}\, du = \frac{u^{5}}{5}$$

            Now substitute $u$ back in:

            $$\frac{\sin^{5}{\left(u \right)}}{5}$$

         So, the result is: $\frac{\sin^{5}{\left(u \right)}}{10}$

      Now substitute $u$ back in:

      $$\frac{\sin^{5}{\left(2 x \right)}}{10}$$

2. Add the constant of integration:

   $$\frac{\sin^{5}{\left(2 x \right)}}{10}+ \mathrm{constant}$$

The answer is:

$$\frac{\sin^{5}{\left(2 x \right)}}{10}+ \mathrm{constant}$$