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

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

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

Then $x=\operatorname{atan}{\left(u \right)}$ and $dx=\left(\operatorname{atan}{\left(u \right)}\right)^{\prime }du = \frac{du}{u^{2} + 1}$ (steps can be seen »).

The integral can be rewritten as

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

Rewrite and split the fraction:

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

Integrate term by term:

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

Apply the constant rule $\int c\, du = c u$ with $c=1$:

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

The integral of $\frac{1}{u^{2} + 1}$ is $\int{\frac{1}{u^{2} + 1} d u} = \operatorname{atan}{\left(u \right)}$:

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

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

$$- \operatorname{atan}{\left({\color{red}{u}} \right)} + {\color{red}{u}} = - \operatorname{atan}{\left({\color{red}{\tan{\left(x \right)}}} \right)} + {\color{red}{\tan{\left(x \right)}}}$$

Therefore,

$$\int{\tan^{2}{\left(x \right)} d x} = \tan{\left(x \right)} - \operatorname{atan}{\left(\tan{\left(x \right)} \right)}$$

Simplify:

$$\int{\tan^{2}{\left(x \right)} d x} = - x + \tan{\left(x \right)}$$

Add the constant of integration:

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

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