Integral of sin2x*cos3xdx 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:

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

      1. Rewrite the integrand:

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

      2. Integrate term-by-term:

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

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

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

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

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

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

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

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

   Method #2

   1. Rewrite the integrand:

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

   2. Integrate term-by-term:

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

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

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

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

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

         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: $2 \cos^{3}{\left(x \right)}$

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

2. Now simplify:

   $$\frac{\cos{\left(x \right)}}{2} - \frac{\cos{\left(5 x \right)}}{10}$$

3. Add the constant of integration:

   $$\frac{\cos{\left(x \right)}}{2} - \frac{\cos{\left(5 x \right)}}{10}+ \mathrm{constant}$$

The answer is:

$$\frac{\cos{\left(x \right)}}{2} - \frac{\cos{\left(5 x \right)}}{10}+ \mathrm{constant}$$