Integral of sin3xsin2x 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:

      $$\int 2 \sin{\left(x \right)} \sin{\left(3 x \right)} \cos{\left(x \right)}\, dx = 2 \int \sin{\left(x \right)} \sin{\left(3 x \right)} \cos{\left(x \right)}\, dx$$

      1. Rewrite the integrand:

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

      2. Integrate term-by-term:

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

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

            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: $- \frac{4 \sin^{5}{\left(x \right)}}{5}$

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

            $$\int 3 \sin^{2}{\left(x \right)} \cos{\left(x \right)}\, dx = 3 \int \sin^{2}{\left(x \right)} \cos{\left(x \right)}\, dx$$

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

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

               $$\int u^{2}\, du$$

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

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

               Now substitute $u$ back in:

               $$\frac{\sin^{3}{\left(x \right)}}{3}$$

            So, the result is: $\sin^{3}{\left(x \right)}$

         The result is: $- \frac{4 \sin^{5}{\left(x \right)}}{5} + \sin^{3}{\left(x \right)}$

      So, the result is: $- \frac{8 \sin^{5}{\left(x \right)}}{5} + 2 \sin^{3}{\left(x \right)}$

   Method #2

   1. Rewrite the integrand:

      $$\sin{\left(2 x \right)} \sin{\left(3 x \right)} = - 8 \sin^{4}{\left(x \right)} \cos{\left(x \right)} + 6 \sin^{2}{\left(x \right)} \cos{\left(x \right)}$$

   2. Integrate term-by-term:

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

         $$\int \left(- 8 \sin^{4}{\left(x \right)} \cos{\left(x \right)}\right)\, dx = - 8 \int \sin^{4}{\left(x \right)} \cos{\left(x \right)}\, dx$$

         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: $- \frac{8 \sin^{5}{\left(x \right)}}{5}$

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

         $$\int 6 \sin^{2}{\left(x \right)} \cos{\left(x \right)}\, dx = 6 \int \sin^{2}{\left(x \right)} \cos{\left(x \right)}\, dx$$

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

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

            $$\int u^{2}\, du$$

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

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

            Now substitute $u$ back in:

            $$\frac{\sin^{3}{\left(x \right)}}{3}$$

         So, the result is: $2 \sin^{3}{\left(x \right)}$

      The result is: $- \frac{8 \sin^{5}{\left(x \right)}}{5} + 2 \sin^{3}{\left(x \right)}$

2. Add the constant of integration:

   $$- \frac{8 \sin^{5}{\left(x \right)}}{5} + 2 \sin^{3}{\left(x \right)}+ \mathrm{constant}$$

The answer is:

$$- \frac{8 \sin^{5}{\left(x \right)}}{5} + 2 \sin^{3}{\left(x \right)}+ \mathrm{constant}$$