Integral of sin(5x)sin(2x)dx 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:

      $$\int 2 \sin{\left(x \right)} \sin{\left(5 x \right)} \cos{\left(x \right)}\, dx = 2 \int \sin{\left(x \right)} \sin{\left(5 x \right)} \cos{\left(x \right)}\, dx$$

      1. Rewrite the integrand:

         $$\sin{\left(x \right)} \sin{\left(5 x \right)} \cos{\left(x \right)} = 16 \sin^{6}{\left(x \right)} \cos{\left(x \right)} - 20 \sin^{4}{\left(x \right)} \cos{\left(x \right)} + 5 \sin^{2}{\left(x \right)} \cos{\left(x \right)}$$

      2. Integrate term-by-term:

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

            $$\int 16 \sin^{6}{\left(x \right)} \cos{\left(x \right)}\, dx = 16 \int \sin^{6}{\left(x \right)} \cos{\left(x \right)}\, dx$$

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

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

               $$\int u^{6}\, du$$

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

                  $$\int u^{6}\, du = \frac{u^{7}}{7}$$

               Now substitute $u$ back in:

               $$\frac{\sin^{7}{\left(x \right)}}{7}$$

            So, the result is: $\frac{16 \sin^{7}{\left(x \right)}}{7}$

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

            $$\int \left(- 20 \sin^{4}{\left(x \right)} \cos{\left(x \right)}\right)\, dx = - 20 \int \sin^{4}{\left(x \right)} \cos{\left(x \right)}\, dx$$

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

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

            $$\int 5 \sin^{2}{\left(x \right)} \cos{\left(x \right)}\, dx = 5 \int \sin^{2}{\left(x \right)} \cos{\left(x \right)}\, dx$$

            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: $\frac{5 \sin^{3}{\left(x \right)}}{3}$

         The result is: $\frac{16 \sin^{7}{\left(x \right)}}{7} - 4 \sin^{5}{\left(x \right)} + \frac{5 \sin^{3}{\left(x \right)}}{3}$

      So, the result is: $\frac{32 \sin^{7}{\left(x \right)}}{7} - 8 \sin^{5}{\left(x \right)} + \frac{10 \sin^{3}{\left(x \right)}}{3}$

   Method #2

   1. Rewrite the integrand:

      $$\sin{\left(5 x \right)} \sin{\left(2 x \right)} 1 = 32 \sin^{6}{\left(x \right)} \cos{\left(x \right)} - 40 \sin^{4}{\left(x \right)} \cos{\left(x \right)} + 10 \sin^{2}{\left(x \right)} \cos{\left(x \right)}$$

   2. Integrate term-by-term:

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

         $$\int 32 \sin^{6}{\left(x \right)} \cos{\left(x \right)}\, dx = 32 \int \sin^{6}{\left(x \right)} \cos{\left(x \right)}\, dx$$

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

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

            $$\int u^{6}\, du$$

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

               $$\int u^{6}\, du = \frac{u^{7}}{7}$$

            Now substitute $u$ back in:

            $$\frac{\sin^{7}{\left(x \right)}}{7}$$

         So, the result is: $\frac{32 \sin^{7}{\left(x \right)}}{7}$

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

         $$\int \left(- 40 \sin^{4}{\left(x \right)} \cos{\left(x \right)}\right)\, dx = - 40 \int \sin^{4}{\left(x \right)} \cos{\left(x \right)}\, dx$$

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

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

         $$\int 10 \sin^{2}{\left(x \right)} \cos{\left(x \right)}\, dx = 10 \int \sin^{2}{\left(x \right)} \cos{\left(x \right)}\, dx$$

         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: $\frac{10 \sin^{3}{\left(x \right)}}{3}$

      The result is: $\frac{32 \sin^{7}{\left(x \right)}}{7} - 8 \sin^{5}{\left(x \right)} + \frac{10 \sin^{3}{\left(x \right)}}{3}$

2. Now simplify:

   $$\frac{2 \cdot \left(48 \sin^{4}{\left(x \right)} - 84 \sin^{2}{\left(x \right)} + 35\right) \sin^{3}{\left(x \right)}}{21}$$

3. Add the constant of integration:

   $$\frac{2 \cdot \left(48 \sin^{4}{\left(x \right)} - 84 \sin^{2}{\left(x \right)} + 35\right) \sin^{3}{\left(x \right)}}{21}+ \mathrm{constant}$$

The answer is: