Integral of $$$\tan^{3}{\left(x \right)} \sec{\left(x \right)}$$$
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Find $$$\int \tan^{3}{\left(x \right)} \sec{\left(x \right)}\, dx$$$.
Solution
Strip out one tangent and write everything else in terms of the secant, using the formula $$$\tan^2\left(x \right)=\sec^2\left(x \right)-1$$$:
$${\color{red}{\int{\tan^{3}{\left(x \right)} \sec{\left(x \right)} d x}}} = {\color{red}{\int{\left(\sec^{2}{\left(x \right)} - 1\right) \tan{\left(x \right)} \sec{\left(x \right)} d x}}}$$
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 becomes
$${\color{red}{\int{\left(\sec^{2}{\left(x \right)} - 1\right) \tan{\left(x \right)} \sec{\left(x \right)} d x}}} = {\color{red}{\int{\left(u^{2} - 1\right)d u}}}$$
Integrate term by term:
$${\color{red}{\int{\left(u^{2} - 1\right)d u}}} = {\color{red}{\left(- \int{1 d u} + \int{u^{2} d u}\right)}}$$
Apply the constant rule $$$\int c\, du = c u$$$ with $$$c=1$$$:
$$\int{u^{2} d u} - {\color{red}{\int{1 d u}}} = \int{u^{2} d u} - {\color{red}{u}}$$
Apply the power rule $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=2$$$:
$$- u + {\color{red}{\int{u^{2} d u}}}=- u + {\color{red}{\frac{u^{1 + 2}}{1 + 2}}}=- u + {\color{red}{\left(\frac{u^{3}}{3}\right)}}$$
Recall that $$$u=\sec{\left(x \right)}$$$:
$$- {\color{red}{u}} + \frac{{\color{red}{u}}^{3}}{3} = - {\color{red}{\sec{\left(x \right)}}} + \frac{{\color{red}{\sec{\left(x \right)}}}^{3}}{3}$$
Therefore,
$$\int{\tan^{3}{\left(x \right)} \sec{\left(x \right)} d x} = \frac{\sec^{3}{\left(x \right)}}{3} - \sec{\left(x \right)}$$
Add the constant of integration:
$$\int{\tan^{3}{\left(x \right)} \sec{\left(x \right)} d x} = \frac{\sec^{3}{\left(x \right)}}{3} - \sec{\left(x \right)}+C$$
Answer
$$$\int \tan^{3}{\left(x \right)} \sec{\left(x \right)}\, dx = \left(\frac{\sec^{3}{\left(x \right)}}{3} - \sec{\left(x \right)}\right) + C$$$A