Integral de $$$\sin{\left(2 x \right)} \cos{\left(2 x \right)}$$$
Calculadora relacionada: Calculadora de integrales definidas e impropias
Tu aportación
Encuentra $$$\int \sin{\left(2 x \right)} \cos{\left(2 x \right)}\, dx$$$.
Solución
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 $$$\cos{\left(2 x \right)} dx = \frac{du}{2}$$$.
Therefore,
$${\color{red}{\int{\sin{\left(2 x \right)} \cos{\left(2 x \right)} d x}}} = {\color{red}{\int{\frac{u}{2} d u}}}$$
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,
$$\int{\sin{\left(2 x \right)} \cos{\left(2 x \right)} d x} = \frac{\sin^{2}{\left(2 x \right)}}{4}$$
Add the constant of integration:
$$\int{\sin{\left(2 x \right)} \cos{\left(2 x \right)} d x} = \frac{\sin^{2}{\left(2 x \right)}}{4}+C$$
Answer: $$$\int{\sin{\left(2 x \right)} \cos{\left(2 x \right)} d x}=\frac{\sin^{2}{\left(2 x \right)}}{4}+C$$$