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

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

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

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

So,

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

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

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

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

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

Therefore,

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

Add the constant of integration:

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

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