Integral of 2*sin(6x)*cos(3x) dx

1. There are multiple ways to do this integral.

   Method #1

   1. Let $u = 3 x$.

      Then let $du = 3 dx$ and substitute $\frac{2 du}{3}$:

      $$\int \frac{2 \sin{\left(2 u \right)} \cos{\left(u \right)}}{3}\, du$$

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

         $$\int \sin{\left(2 u \right)} \cos{\left(u \right)}\, du = \frac{2 \int \sin{\left(2 u \right)} \cos{\left(u \right)}\, du}{3}$$

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

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

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

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

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

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

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

                  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:

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

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

         So, the result is: $- \frac{4 \cos^{3}{\left(u \right)}}{9}$

      Now substitute $u$ back in:

      $$- \frac{4 \cos^{3}{\left(3 x \right)}}{9}$$

   Method #2

   1. Rewrite the integrand:

      $$2 \sin{\left(6 x \right)} \cos{\left(3 x \right)} = 256 \sin^{5}{\left(x \right)} \cos^{4}{\left(x \right)} - 192 \sin^{5}{\left(x \right)} \cos^{2}{\left(x \right)} - 256 \sin^{3}{\left(x \right)} \cos^{4}{\left(x \right)} + 192 \sin^{3}{\left(x \right)} \cos^{2}{\left(x \right)} + 48 \sin{\left(x \right)} \cos^{4}{\left(x \right)} - 36 \sin{\left(x \right)} \cos^{2}{\left(x \right)}$$

   2. Integrate term-by-term:

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

         $$\int 256 \sin^{5}{\left(x \right)} \cos^{4}{\left(x \right)}\, dx = 256 \int \sin^{5}{\left(x \right)} \cos^{4}{\left(x \right)}\, dx$$

         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. 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:

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

         So, the result is: $- \frac{256 \cos^{9}{\left(x \right)}}{9} + \frac{512 \cos^{7}{\left(x \right)}}{7} - \frac{256 \cos^{5}{\left(x \right)}}{5}$

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

         $$\int \left(- 192 \sin^{5}{\left(x \right)} \cos^{2}{\left(x \right)}\right)\, dx = - 192 \int \sin^{5}{\left(x \right)} \cos^{2}{\left(x \right)}\, dx$$

         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: $\frac{192 \cos^{7}{\left(x \right)}}{7} - \frac{384 \cos^{5}{\left(x \right)}}{5} + 64 \cos^{3}{\left(x \right)}$

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

         $$\int \left(- 256 \sin^{3}{\left(x \right)} \cos^{4}{\left(x \right)}\right)\, dx = - 256 \int \sin^{3}{\left(x \right)} \cos^{4}{\left(x \right)}\, dx$$

         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: $- \frac{256 \cos^{7}{\left(x \right)}}{7} + \frac{256 \cos^{5}{\left(x \right)}}{5}$

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

         $$\int 192 \sin^{3}{\left(x \right)} \cos^{2}{\left(x \right)}\, dx = 192 \int \sin^{3}{\left(x \right)} \cos^{2}{\left(x \right)}\, dx$$

         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: $\frac{192 \cos^{5}{\left(x \right)}}{5} - 64 \cos^{3}{\left(x \right)}$

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

         $$\int 48 \sin{\left(x \right)} \cos^{4}{\left(x \right)}\, dx = 48 \int \sin{\left(x \right)} \cos^{4}{\left(x \right)}\, dx$$

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

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

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

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

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

               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: $- \frac{48 \cos^{5}{\left(x \right)}}{5}$

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

         $$\int \left(- 36 \sin{\left(x \right)} \cos^{2}{\left(x \right)}\right)\, dx = - 36 \int \sin{\left(x \right)} \cos^{2}{\left(x \right)}\, dx$$

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

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

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

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

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

               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: $12 \cos^{3}{\left(x \right)}$

      The result is: $- \frac{256 \cos^{9}{\left(x \right)}}{9} + 64 \cos^{7}{\left(x \right)} - 48 \cos^{5}{\left(x \right)} + 12 \cos^{3}{\left(x \right)}$

2. Add the constant of integration:

   $$- \frac{4 \cos^{3}{\left(3 x \right)}}{9}+ \mathrm{constant}$$

The answer is:

$$- \frac{4 \cos^{3}{\left(3 x \right)}}{9}+ \mathrm{constant}$$