Integral of sin(2x)dx dx

1. There are multiple ways to do this integral.

   Method #1

   1. Let $u = 2 x$.

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

      $$\int \frac{\sin{\left(u \right)}}{4}\, du$$

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

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

         1. The integral of sine is negative cosine:

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

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

      Now substitute $u$ back in:

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

   Method #2

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

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

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

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

         $$\int u\, du$$

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

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

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

2. Add the constant of integration:

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

The answer is:

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