Integral of $$$\tan^{3}{\left(x \right)} \sec^{6}{\left(x \right)}$$$

Find $$$\int \tan^{3}{\left(x \right)} \sec^{6}{\left(x \right)}\, dx$$$.

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

$${\color{red}{\int{\tan^{3}{\left(x \right)} \sec^{6}{\left(x \right)} d x}}} = {\color{red}{\int{\left(\sec^{2}{\left(x \right)} - 1\right) \tan{\left(x \right)} \sec^{6}{\left(x \right)} d x}}}$$

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

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

The integral can be rewritten as

$${\color{red}{\int{\left(\sec^{2}{\left(x \right)} - 1\right) \tan{\left(x \right)} \sec^{6}{\left(x \right)} d x}}} = {\color{red}{\int{u^{5} \left(u^{2} - 1\right) d u}}}$$

Expand the expression:

$${\color{red}{\int{u^{5} \left(u^{2} - 1\right) d u}}} = {\color{red}{\int{\left(u^{7} - u^{5}\right)d u}}}$$

Integrate term by term:

$${\color{red}{\int{\left(u^{7} - u^{5}\right)d u}}} = {\color{red}{\left(- \int{u^{5} d u} + \int{u^{7} d u}\right)}}$$

Apply the power rule $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=7$$$:

$$- \int{u^{5} d u} + {\color{red}{\int{u^{7} d u}}}=- \int{u^{5} d u} + {\color{red}{\frac{u^{1 + 7}}{1 + 7}}}=- \int{u^{5} d u} + {\color{red}{\left(\frac{u^{8}}{8}\right)}}$$

Apply the power rule $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=5$$$:

$$\frac{u^{8}}{8} - {\color{red}{\int{u^{5} d u}}}=\frac{u^{8}}{8} - {\color{red}{\frac{u^{1 + 5}}{1 + 5}}}=\frac{u^{8}}{8} - {\color{red}{\left(\frac{u^{6}}{6}\right)}}$$

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

$$- \frac{{\color{red}{u}}^{6}}{6} + \frac{{\color{red}{u}}^{8}}{8} = - \frac{{\color{red}{\sec{\left(x \right)}}}^{6}}{6} + \frac{{\color{red}{\sec{\left(x \right)}}}^{8}}{8}$$

Therefore,

$$\int{\tan^{3}{\left(x \right)} \sec^{6}{\left(x \right)} d x} = \frac{\sec^{8}{\left(x \right)}}{8} - \frac{\sec^{6}{\left(x \right)}}{6}$$

Add the constant of integration:

$$\int{\tan^{3}{\left(x \right)} \sec^{6}{\left(x \right)} d x} = \frac{\sec^{8}{\left(x \right)}}{8} - \frac{\sec^{6}{\left(x \right)}}{6}+C$$

$$$\int \tan^{3}{\left(x \right)} \sec^{6}{\left(x \right)}\, dx = \left(\frac{\sec^{8}{\left(x \right)}}{8} - \frac{\sec^{6}{\left(x \right)}}{6}\right) + C$$$A