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

The calculator will find the integral/antiderivative of $\sin{\left(2 x \right)} \cos{\left(2 x \right)}$, with steps shown.

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

### Solution

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

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

So,

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

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 $f{\left(u \right)} = u$:

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

Apply the power rule $\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$ $\left(n \neq -1 \right)$ with $n=1$:

$$\frac{{\color{red}{\int{u d u}}}}{2}=\frac{{\color{red}{\frac{u^{1 + 1}}{1 + 1}}}}{2}=\frac{{\color{red}{\left(\frac{u^{2}}{2}\right)}}}{2}$$

Recall that $u=\sin{\left(2 x \right)}$:

$$\frac{{\color{red}{u}}^{2}}{4} = \frac{{\color{red}{\sin{\left(2 x \right)}}}^{2}}{4}$$

Therefore,

$$\int{\sin{\left(2 x \right)} \cos{\left(2 x \right)} d x} = \frac{\sin^{2}{\left(2 x \right)}}{4}$$

$$\int{\sin{\left(2 x \right)} \cos{\left(2 x \right)} d x} = \frac{\sin^{2}{\left(2 x \right)}}{4}+C$$
Answer: $\int{\sin{\left(2 x \right)} \cos{\left(2 x \right)} d x}=\frac{\sin^{2}{\left(2 x \right)}}{4}+C$