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

1. Rewrite the integrand:

   $$\sin{\left(5 x \right)} \cos{\left(4 x \right)} = 128 \sin^{5}{\left(x \right)} \cos^{4}{\left(x \right)} - 128 \sin^{5}{\left(x \right)} \cos^{2}{\left(x \right)} + 16 \sin^{5}{\left(x \right)} - 160 \sin^{3}{\left(x \right)} \cos^{4}{\left(x \right)} + 160 \sin^{3}{\left(x \right)} \cos^{2}{\left(x \right)} - 20 \sin^{3}{\left(x \right)} + 40 \sin{\left(x \right)} \cos^{4}{\left(x \right)} - 40 \sin{\left(x \right)} \cos^{2}{\left(x \right)} + 5 \sin{\left(x \right)}$$

2. Integrate term-by-term:

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

      $$\int 128 \sin^{5}{\left(x \right)} \cos^{4}{\left(x \right)}\, dx = 128 \int \sin^{5}{\left(x \right)} \cos^{4}{\left(x \right)}\, dx$$

      1. Rewrite the integrand:

         $$\sin^{5}{\left(x \right)} \cos^{4}{\left(x \right)} = \left(1 - \cos^{2}{\left(x \right)}\right)^{2} \sin{\left(x \right)} \cos^{4}{\left(x \right)}$$

      2. There are multiple ways to do this integral.

         Method #1

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

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

            $$\int \left(- u^{8} + 2 u^{6} - u^{4}\right)\, du$$

            1. Integrate term-by-term:

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

                  $$\int \left(- u^{8}\right)\, du = - \int u^{8}\, du$$

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

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

                  So, the result is: $- \frac{u^{9}}{9}$

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

                  $$\int 2 u^{6}\, du = 2 \int u^{6}\, du$$

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

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

                  So, the result is: $\frac{2 u^{7}}{7}$

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

                  $$\int \left(- u^{4}\right)\, du = - \int u^{4}\, du$$

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

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

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

               The result is: $- \frac{u^{9}}{9} + \frac{2 u^{7}}{7} - \frac{u^{5}}{5}$

            Now substitute $u$ back in:

            $$- \frac{\cos^{9}{\left(x \right)}}{9} + \frac{2 \cos^{7}{\left(x \right)}}{7} - \frac{\cos^{5}{\left(x \right)}}{5}$$

         Method #2

         1. Rewrite the integrand:

            $$\left(1 - \cos^{2}{\left(x \right)}\right)^{2} \sin{\left(x \right)} \cos^{4}{\left(x \right)} = \sin{\left(x \right)} \cos^{8}{\left(x \right)} - 2 \sin{\left(x \right)} \cos^{6}{\left(x \right)} + \sin{\left(x \right)} \cos^{4}{\left(x \right)}$$

         2. Integrate term-by-term:

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

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

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

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

                  $$\int \left(- u^{8}\right)\, du = - \int u^{8}\, du$$

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

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

                  So, the result is: $- \frac{u^{9}}{9}$

               Now substitute $u$ back in:

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

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

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

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

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

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

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

                     $$\int \left(- u^{6}\right)\, du = - \int u^{6}\, du$$

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

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

                     So, the result is: $- \frac{u^{7}}{7}$

                  Now substitute $u$ back in:

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

               So, the result is: $\frac{2 \cos^{7}{\left(x \right)}}{7}$

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

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

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

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

                  $$\int \left(- u^{4}\right)\, du = - \int u^{4}\, du$$

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

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

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

               Now substitute $u$ back in:

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

            The result is: $- \frac{\cos^{9}{\left(x \right)}}{9} + \frac{2 \cos^{7}{\left(x \right)}}{7} - \frac{\cos^{5}{\left(x \right)}}{5}$

         Method #3

         1. Rewrite the integrand:

            $$\left(1 - \cos^{2}{\left(x \right)}\right)^{2} \sin{\left(x \right)} \cos^{4}{\left(x \right)} = \sin{\left(x \right)} \cos^{8}{\left(x \right)} - 2 \sin{\left(x \right)} \cos^{6}{\left(x \right)} + \sin{\left(x \right)} \cos^{4}{\left(x \right)}$$

         2. Integrate term-by-term:

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

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

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

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

                  $$\int \left(- u^{8}\right)\, du = - \int u^{8}\, du$$

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

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

                  So, the result is: $- \frac{u^{9}}{9}$

               Now substitute $u$ back in:

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

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

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

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

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

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

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

                     $$\int \left(- u^{6}\right)\, du = - \int u^{6}\, du$$

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

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

                     So, the result is: $- \frac{u^{7}}{7}$

                  Now substitute $u$ back in:

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

               So, the result is: $\frac{2 \cos^{7}{\left(x \right)}}{7}$

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

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

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

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

                  $$\int \left(- u^{4}\right)\, du = - \int u^{4}\, du$$

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

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

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

               Now substitute $u$ back in:

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

            The result is: $- \frac{\cos^{9}{\left(x \right)}}{9} + \frac{2 \cos^{7}{\left(x \right)}}{7} - \frac{\cos^{5}{\left(x \right)}}{5}$

      So, the result is: $- \frac{128 \cos^{9}{\left(x \right)}}{9} + \frac{256 \cos^{7}{\left(x \right)}}{7} - \frac{128 \cos^{5}{\left(x \right)}}{5}$

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

      $$\int \left(- 128 \sin^{5}{\left(x \right)} \cos^{2}{\left(x \right)}\right)\, dx = - 128 \int \sin^{5}{\left(x \right)} \cos^{2}{\left(x \right)}\, dx$$

      1. Rewrite the integrand:

         $$\sin^{5}{\left(x \right)} \cos^{2}{\left(x \right)} = \left(1 - \cos^{2}{\left(x \right)}\right)^{2} \sin{\left(x \right)} \cos^{2}{\left(x \right)}$$

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

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

         $$\int \left(- u^{6} + 2 u^{4} - u^{2}\right)\, du$$

         1. Integrate term-by-term:

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

               $$\int \left(- u^{6}\right)\, du = - \int u^{6}\, du$$

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

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

               So, the result is: $- \frac{u^{7}}{7}$

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

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

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

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

               So, the result is: $\frac{2 u^{5}}{5}$

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

               $$\int \left(- u^{2}\right)\, du = - \int u^{2}\, du$$

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

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

               So, the result is: $- \frac{u^{3}}{3}$

            The result is: $- \frac{u^{7}}{7} + \frac{2 u^{5}}{5} - \frac{u^{3}}{3}$

         Now substitute $u$ back in:

         $$- \frac{\cos^{7}{\left(x \right)}}{7} + \frac{2 \cos^{5}{\left(x \right)}}{5} - \frac{\cos^{3}{\left(x \right)}}{3}$$

      So, the result is: $\frac{128 \cos^{7}{\left(x \right)}}{7} - \frac{256 \cos^{5}{\left(x \right)}}{5} + \frac{128 \cos^{3}{\left(x \right)}}{3}$

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

      $$\int 16 \sin^{5}{\left(x \right)}\, dx = 16 \int \sin^{5}{\left(x \right)}\, dx$$

      1. Rewrite the integrand:

         $$\sin^{5}{\left(x \right)} = \left(1 - \cos^{2}{\left(x \right)}\right)^{2} \sin{\left(x \right)}$$

      2. Rewrite the integrand:

         $$\left(1 - \cos^{2}{\left(x \right)}\right)^{2} \sin{\left(x \right)} = \sin{\left(x \right)} \cos^{4}{\left(x \right)} - 2 \sin{\left(x \right)} \cos^{2}{\left(x \right)} + \sin{\left(x \right)}$$

      3. Integrate term-by-term:

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

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

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

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

               $$\int \left(- u^{4}\right)\, du = - \int u^{4}\, du$$

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

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

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

            Now substitute $u$ back in:

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

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

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

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

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

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

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

                  $$\int \left(- u^{2}\right)\, du = - \int u^{2}\, du$$

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

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

                  So, the result is: $- \frac{u^{3}}{3}$

               Now substitute $u$ back in:

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

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

         1. The integral of sine is negative cosine:

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

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

      So, the result is: $- \frac{16 \cos^{5}{\left(x \right)}}{5} + \frac{32 \cos^{3}{\left(x \right)}}{3} - 16 \cos{\left(x \right)}$

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

      $$\int \left(- 160 \sin^{3}{\left(x \right)} \cos^{4}{\left(x \right)}\right)\, dx = - 160 \int \sin^{3}{\left(x \right)} \cos^{4}{\left(x \right)}\, dx$$

      1. Rewrite the integrand:

         $$\sin^{3}{\left(x \right)} \cos^{4}{\left(x \right)} = \left(1 - \cos^{2}{\left(x \right)}\right) \sin{\left(x \right)} \cos^{4}{\left(x \right)}$$

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

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

         $$\int \left(u^{6} - u^{4}\right)\, du$$

         1. Integrate term-by-term:

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

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

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

               $$\int \left(- u^{4}\right)\, du = - \int u^{4}\, du$$

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

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

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

            The result is: $\frac{u^{7}}{7} - \frac{u^{5}}{5}$

         Now substitute $u$ back in:

         $$\frac{\cos^{7}{\left(x \right)}}{7} - \frac{\cos^{5}{\left(x \right)}}{5}$$

      So, the result is: $- \frac{160 \cos^{7}{\left(x \right)}}{7} + 32 \cos^{5}{\left(x \right)}$

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

      $$\int 160 \sin^{3}{\left(x \right)} \cos^{2}{\left(x \right)}\, dx = 160 \int \sin^{3}{\left(x \right)} \cos^{2}{\left(x \right)}\, dx$$

      1. Rewrite the integrand:

         $$\sin^{3}{\left(x \right)} \cos^{2}{\left(x \right)} = \left(1 - \cos^{2}{\left(x \right)}\right) \sin{\left(x \right)} \cos^{2}{\left(x \right)}$$

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

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

         $$\int \left(u^{4} - u^{2}\right)\, du$$

         1. Integrate term-by-term:

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

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

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

               $$\int \left(- u^{2}\right)\, du = - \int u^{2}\, du$$

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

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

               So, the result is: $- \frac{u^{3}}{3}$

            The result is: $\frac{u^{5}}{5} - \frac{u^{3}}{3}$

         Now substitute $u$ back in:

         $$\frac{\cos^{5}{\left(x \right)}}{5} - \frac{\cos^{3}{\left(x \right)}}{3}$$

      So, the result is: $32 \cos^{5}{\left(x \right)} - \frac{160 \cos^{3}{\left(x \right)}}{3}$

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

      $$\int \left(- 20 \sin^{3}{\left(x \right)}\right)\, dx = - 20 \int \sin^{3}{\left(x \right)}\, dx$$

      1. Rewrite the integrand:

         $$\sin^{3}{\left(x \right)} = \left(1 - \cos^{2}{\left(x \right)}\right) \sin{\left(x \right)}$$

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

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

         $$\int \left(u^{2} - 1\right)\, du$$

         1. Integrate term-by-term:

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

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

            1. The integral of a constant is the constant times the variable of integration:

               $$\int \left(-1\right)\, du = - u$$

            The result is: $\frac{u^{3}}{3} - u$

         Now substitute $u$ back in:

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

      So, the result is: $- \frac{20 \cos^{3}{\left(x \right)}}{3} + 20 \cos{\left(x \right)}$

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

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

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

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

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

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

            $$\int \left(- u^{4}\right)\, du = - \int u^{4}\, du$$

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

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

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

         Now substitute $u$ back in:

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

      So, the result is: $- 8 \cos^{5}{\left(x \right)}$

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

      $$\int \left(- 40 \sin{\left(x \right)} \cos^{2}{\left(x \right)}\right)\, dx = - 40 \int \sin{\left(x \right)} \cos^{2}{\left(x \right)}\, dx$$

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

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

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

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

            $$\int \left(- u^{2}\right)\, du = - \int u^{2}\, du$$

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

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

            So, the result is: $- \frac{u^{3}}{3}$

         Now substitute $u$ back in:

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

      So, the result is: $\frac{40 \cos^{3}{\left(x \right)}}{3}$

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

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

      1. The integral of sine is negative cosine:

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

      So, the result is: $- 5 \cos{\left(x \right)}$

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

3. Now simplify:

   $$\frac{\left(- 128 \cos^{8}{\left(x \right)} + 288 \cos^{6}{\left(x \right)} - 216 \cos^{4}{\left(x \right)} + 60 \cos^{2}{\left(x \right)} - 9\right) \cos{\left(x \right)}}{9}$$

4. Add the constant of integration:

   $$\frac{\left(- 128 \cos^{8}{\left(x \right)} + 288 \cos^{6}{\left(x \right)} - 216 \cos^{4}{\left(x \right)} + 60 \cos^{2}{\left(x \right)} - 9\right) \cos{\left(x \right)}}{9}+ \mathrm{constant}$$

The answer is:

$$\frac{\left(- 128 \cos^{8}{\left(x \right)} + 288 \cos^{6}{\left(x \right)} - 216 \cos^{4}{\left(x \right)} + 60 \cos^{2}{\left(x \right)} - 9\right) \cos{\left(x \right)}}{9}+ \mathrm{constant}$$