Integral of $$$\tan^{3}{\left(2 x \right)} \sec{\left(2 x \right)}$$$

Find $$$\int \tan^{3}{\left(2 x \right)} \sec{\left(2 x \right)}\, dx$$$.

Let $$$u=2 x$$$.

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

The integral becomes

$${\color{red}{\int{\tan^{3}{\left(2 x \right)} \sec{\left(2 x \right)} d x}}} = {\color{red}{\int{\frac{\tan^{3}{\left(u \right)} \sec{\left(u \right)}}{2} d u}}}$$

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 $$$f{\left(u \right)} = \tan^{3}{\left(u \right)} \sec{\left(u \right)}$$$:

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

Strip out one tangent and write everything else in terms of the secant, using the formula $$$\tan^2\left( u \right)=\sec^2\left( u \right)-1$$$:

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

Let $$$v=\sec{\left(u \right)}$$$.

Then $$$dv=\left(\sec{\left(u \right)}\right)^{\prime }du = \tan{\left(u \right)} \sec{\left(u \right)} du$$$ (steps can be seen »), and we have that $$$\tan{\left(u \right)} \sec{\left(u \right)} du = dv$$$.

Thus,

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

Integrate term by term:

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

Apply the constant rule $$$\int c\, dv = c v$$$ with $$$c=1$$$:

$$\frac{\int{v^{2} d v}}{2} - \frac{{\color{red}{\int{1 d v}}}}{2} = \frac{\int{v^{2} d v}}{2} - \frac{{\color{red}{v}}}{2}$$

Apply the power rule $$$\int v^{n}\, dv = \frac{v^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=2$$$:

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

Recall that $$$v=\sec{\left(u \right)}$$$:

$$- \frac{{\color{red}{v}}}{2} + \frac{{\color{red}{v}}^{3}}{6} = - \frac{{\color{red}{\sec{\left(u \right)}}}}{2} + \frac{{\color{red}{\sec{\left(u \right)}}}^{3}}{6}$$

Recall that $$$u=2 x$$$:

$$- \frac{\sec{\left({\color{red}{u}} \right)}}{2} + \frac{\sec^{3}{\left({\color{red}{u}} \right)}}{6} = - \frac{\sec{\left({\color{red}{\left(2 x\right)}} \right)}}{2} + \frac{\sec^{3}{\left({\color{red}{\left(2 x\right)}} \right)}}{6}$$

Therefore,

$$\int{\tan^{3}{\left(2 x \right)} \sec{\left(2 x \right)} d x} = \frac{\sec^{3}{\left(2 x \right)}}{6} - \frac{\sec{\left(2 x \right)}}{2}$$

Simplify:

$$\int{\tan^{3}{\left(2 x \right)} \sec{\left(2 x \right)} d x} = \frac{\left(\sec^{2}{\left(2 x \right)} - 3\right) \sec{\left(2 x \right)}}{6}$$

Add the constant of integration:

$$\int{\tan^{3}{\left(2 x \right)} \sec{\left(2 x \right)} d x} = \frac{\left(\sec^{2}{\left(2 x \right)} - 3\right) \sec{\left(2 x \right)}}{6}+C$$

$$$\int \tan^{3}{\left(2 x \right)} \sec{\left(2 x \right)}\, dx = \frac{\left(\sec^{2}{\left(2 x \right)} - 3\right) \sec{\left(2 x \right)}}{6} + C$$$A