Integral of sin^3x*sin2x dx

1. There are multiple ways to do this integral.

   Method #1

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

      $$\int 2 \sin^{4}{\left(x \right)} \cos{\left(x \right)}\, dx = 2 \int \sin^{4}{\left(x \right)} \cos{\left(x \right)}\, dx$$

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

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

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

      So, the result is: $\frac{2 \sin^{5}{\left(x \right)}}{5}$

   Method #2

   1. Rewrite the integrand:

      $$\sin^{3}{\left(x \right)} \sin{\left(2 x \right)} = 2 \sin^{4}{\left(x \right)} \cos{\left(x \right)}$$

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

      $$\int 2 \sin^{4}{\left(x \right)} \cos{\left(x \right)}\, dx = 2 \int \sin^{4}{\left(x \right)} \cos{\left(x \right)}\, dx$$

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

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

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

      So, the result is: $\frac{2 \sin^{5}{\left(x \right)}}{5}$

2. Add the constant of integration:

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

The answer is:

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