Integral of $$$\frac{\sin{\left(2 x \right)}}{\cos^{3}{\left(2 x \right)}}$$$

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

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

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

Therefore,

$${\color{red}{\int{\frac{\sin{\left(2 x \right)}}{\cos^{3}{\left(2 x \right)}} d x}}} = {\color{red}{\int{\left(- \frac{1}{2 u^{3}}\right)d u}}}$$

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 $$$f{\left(u \right)} = \frac{1}{u^{3}}$$$:

$${\color{red}{\int{\left(- \frac{1}{2 u^{3}}\right)d u}}} = {\color{red}{\left(- \frac{\int{\frac{1}{u^{3}} d u}}{2}\right)}}$$

Apply the power rule $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=-3$$$:

$$- \frac{{\color{red}{\int{\frac{1}{u^{3}} d u}}}}{2}=- \frac{{\color{red}{\int{u^{-3} d u}}}}{2}=- \frac{{\color{red}{\frac{u^{-3 + 1}}{-3 + 1}}}}{2}=- \frac{{\color{red}{\left(- \frac{u^{-2}}{2}\right)}}}{2}=- \frac{{\color{red}{\left(- \frac{1}{2 u^{2}}\right)}}}{2}$$

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

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

Therefore,

$$\int{\frac{\sin{\left(2 x \right)}}{\cos^{3}{\left(2 x \right)}} d x} = \frac{1}{4 \cos^{2}{\left(2 x \right)}}$$

Add the constant of integration:

$$\int{\frac{\sin{\left(2 x \right)}}{\cos^{3}{\left(2 x \right)}} d x} = \frac{1}{4 \cos^{2}{\left(2 x \right)}}+C$$

$$$\int \frac{\sin{\left(2 x \right)}}{\cos^{3}{\left(2 x \right)}}\, dx = \frac{1}{4 \cos^{2}{\left(2 x \right)}} + C$$$A