Integral of $\sin{\left(2 x \right)}$

Find $\int \sin{\left(2 x \right)}\, dx$.

Let $u=2 x$.

Then $du=\left(2 x\right)^{\prime }dx = 2 dx$ (steps can be seen »), and we have that $dx = \frac{du}{2}$.

So,

$${\color{red}{\int{\sin{\left(2 x \right)} d x}}} = {\color{red}{\int{\frac{\sin{\left(u \right)}}{2} d u}}}$$

Apply the constant multiple rule $\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$ with $c=\frac{1}{2}$ and $f{\left(u \right)} = \sin{\left(u \right)}$:

$${\color{red}{\int{\frac{\sin{\left(u \right)}}{2} d u}}} = {\color{red}{\left(\frac{\int{\sin{\left(u \right)} d u}}{2}\right)}}$$

The integral of the sine is $\int{\sin{\left(u \right)} d u} = - \cos{\left(u \right)}$:

$$\frac{{\color{red}{\int{\sin{\left(u \right)} d u}}}}{2} = \frac{{\color{red}{\left(- \cos{\left(u \right)}\right)}}}{2}$$

Recall that $u=2 x$:

$$- \frac{\cos{\left({\color{red}{u}} \right)}}{2} = - \frac{\cos{\left({\color{red}{\left(2 x\right)}} \right)}}{2}$$

Therefore,

$$\int{\sin{\left(2 x \right)} d x} = - \frac{\cos{\left(2 x \right)}}{2}$$

Add the constant of integration:

$$\int{\sin{\left(2 x \right)} d x} = - \frac{\cos{\left(2 x \right)}}{2}+C$$

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