Integral of sin(4x)*sin(5x) dx

1. Rewrite the integrand:

   $$\sin{\left(4 x \right)} \sin{\left(5 x \right)} = - 128 \sin^{8}{\left(x \right)} \cos{\left(x \right)} + 224 \sin^{6}{\left(x \right)} \cos{\left(x \right)} - 120 \sin^{4}{\left(x \right)} \cos{\left(x \right)} + 20 \sin^{2}{\left(x \right)} \cos{\left(x \right)}$$

2. Integrate term-by-term:

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

      $$\int \left(- 128 \sin^{8}{\left(x \right)} \cos{\left(x \right)}\right)\, dx = - 128 \int \sin^{8}{\left(x \right)} \cos{\left(x \right)}\, dx$$

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

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

         $$\int u^{8}\, du$$

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

            $$\int u^{8}\, du = \frac{u^{9}}{9}$$

         Now substitute $u$ back in:

         $$\frac{\sin^{9}{\left(x \right)}}{9}$$

      So, the result is: $- \frac{128 \sin^{9}{\left(x \right)}}{9}$

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

      $$\int 224 \sin^{6}{\left(x \right)} \cos{\left(x \right)}\, dx = 224 \int \sin^{6}{\left(x \right)} \cos{\left(x \right)}\, dx$$

      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: $32 \sin^{7}{\left(x \right)}$

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

      $$\int \left(- 120 \sin^{4}{\left(x \right)} \cos{\left(x \right)}\right)\, dx = - 120 \int \sin^{4}{\left(x \right)} \cos{\left(x \right)}\, dx$$

      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: $- 24 \sin^{5}{\left(x \right)}$

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

      $$\int 20 \sin^{2}{\left(x \right)} \cos{\left(x \right)}\, dx = 20 \int \sin^{2}{\left(x \right)} \cos{\left(x \right)}\, dx$$

      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: $\frac{20 \sin^{3}{\left(x \right)}}{3}$

   The result is: $- \frac{128 \sin^{9}{\left(x \right)}}{9} + 32 \sin^{7}{\left(x \right)} - 24 \sin^{5}{\left(x \right)} + \frac{20 \sin^{3}{\left(x \right)}}{3}$

3. Now simplify:

   $$\frac{4 \left(- 32 \sin^{6}{\left(x \right)} + 72 \sin^{4}{\left(x \right)} - 54 \sin^{2}{\left(x \right)} + 15\right) \sin^{3}{\left(x \right)}}{9}$$

4. Add the constant of integration:

   $$\frac{4 \left(- 32 \sin^{6}{\left(x \right)} + 72 \sin^{4}{\left(x \right)} - 54 \sin^{2}{\left(x \right)} + 15\right) \sin^{3}{\left(x \right)}}{9}+ \mathrm{constant}$$

The answer is:

$$\frac{4 \left(- 32 \sin^{6}{\left(x \right)} + 72 \sin^{4}{\left(x \right)} - 54 \sin^{2}{\left(x \right)} + 15\right) \sin^{3}{\left(x \right)}}{9}+ \mathrm{constant}$$