Integral of sin(2*x)*cos(2*x)/2 dx

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

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

   1. There are multiple ways to do this integral.

      Method #1

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

         Then let $du = - 2 \sin{\left(2 x \right)} dx$ and substitute $- \frac{du}{2}$:

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

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

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

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

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

            So, the result is: $- \frac{u^{2}}{4}$

         Now substitute $u$ back in:

         $$- \frac{\cos^{2}{\left(2 x \right)}}{4}$$

      Method #2

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

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

         1. Rewrite the integrand:

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

         2. Integrate term-by-term:

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

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

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

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

                  $$\int \left(- u^{3}\right)\, du$$

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

                     $$\int u^{3}\, du = - \int u^{3}\, du$$

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

                        $$\int u^{3}\, du = \frac{u^{4}}{4}$$

                     So, the result is: $- \frac{u^{4}}{4}$

                  Now substitute $u$ back in:

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

               So, the result is: $- \frac{\cos^{4}{\left(x \right)}}{2}$

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

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

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

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

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

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

                     $$\int u\, du = - \int u\, du$$

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

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

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

                  Now substitute $u$ back in:

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

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

            The result is: $- \frac{\cos^{4}{\left(x \right)}}{2} + \frac{\cos^{2}{\left(x \right)}}{2}$

         So, the result is: $- \cos^{4}{\left(x \right)} + \cos^{2}{\left(x \right)}$

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

2. Add the constant of integration:

   $$- \frac{\cos^{2}{\left(2 x \right)}}{8}+ \mathrm{constant}$$

The answer is:

$$- \frac{\cos^{2}{\left(2 x \right)}}{8}+ \mathrm{constant}$$