Integral of $\tan^{7}{\left(x \right)}$

Find $\int \tan^{7}{\left(x \right)}\, dx$.

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

$${\color{red}{\int{\tan^{7}{\left(x \right)} d x}}} = {\color{red}{\int{\frac{u^{7}}{u^{2} + 1} d u}}}$$

Since the degree of the numerator is not less than the degree of the denominator, perform polynomial long division (steps can be seen »):

$${\color{red}{\int{\frac{u^{7}}{u^{2} + 1} d u}}} = {\color{red}{\int{\left(u^{5} - u^{3} + u - \frac{u}{u^{2} + 1}\right)d u}}}$$

Integrate term by term:

$${\color{red}{\int{\left(u^{5} - u^{3} + u - \frac{u}{u^{2} + 1}\right)d u}}} = {\color{red}{\left(\int{u d u} - \int{u^{3} d u} + \int{u^{5} d u} - \int{\frac{u}{u^{2} + 1} d u}\right)}}$$

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

$$- \int{u^{3} d u} + \int{u^{5} d u} - \int{\frac{u}{u^{2} + 1} d u} + {\color{red}{\int{u d u}}}=- \int{u^{3} d u} + \int{u^{5} d u} - \int{\frac{u}{u^{2} + 1} d u} + {\color{red}{\frac{u^{1 + 1}}{1 + 1}}}=- \int{u^{3} d u} + \int{u^{5} d u} - \int{\frac{u}{u^{2} + 1} d u} + {\color{red}{\left(\frac{u^{2}}{2}\right)}}$$

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

$$\frac{u^{2}}{2} - \int{u^{3} d u} - \int{\frac{u}{u^{2} + 1} d u} + {\color{red}{\int{u^{5} d u}}}=\frac{u^{2}}{2} - \int{u^{3} d u} - \int{\frac{u}{u^{2} + 1} d u} + {\color{red}{\frac{u^{1 + 5}}{1 + 5}}}=\frac{u^{2}}{2} - \int{u^{3} d u} - \int{\frac{u}{u^{2} + 1} d u} + {\color{red}{\left(\frac{u^{6}}{6}\right)}}$$

Apply the power rule $\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$ $\left(n \neq -1 \right)$ with $n=3$:

$$\frac{u^{6}}{6} + \frac{u^{2}}{2} - \int{\frac{u}{u^{2} + 1} d u} - {\color{red}{\int{u^{3} d u}}}=\frac{u^{6}}{6} + \frac{u^{2}}{2} - \int{\frac{u}{u^{2} + 1} d u} - {\color{red}{\frac{u^{1 + 3}}{1 + 3}}}=\frac{u^{6}}{6} + \frac{u^{2}}{2} - \int{\frac{u}{u^{2} + 1} d u} - {\color{red}{\left(\frac{u^{4}}{4}\right)}}$$

Let $v=u^{2} + 1$.

Then $dv=\left(u^{2} + 1\right)^{\prime }du = 2 u du$ (steps can be seen »), and we have that $u du = \frac{dv}{2}$.

Therefore,

$$\frac{u^{6}}{6} - \frac{u^{4}}{4} + \frac{u^{2}}{2} - {\color{red}{\int{\frac{u}{u^{2} + 1} d u}}} = \frac{u^{6}}{6} - \frac{u^{4}}{4} + \frac{u^{2}}{2} - {\color{red}{\int{\frac{1}{2 v} d v}}}$$

Apply the constant multiple rule $\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$ with $c=\frac{1}{2}$ and $f{\left(v \right)} = \frac{1}{v}$:

$$\frac{u^{6}}{6} - \frac{u^{4}}{4} + \frac{u^{2}}{2} - {\color{red}{\int{\frac{1}{2 v} d v}}} = \frac{u^{6}}{6} - \frac{u^{4}}{4} + \frac{u^{2}}{2} - {\color{red}{\left(\frac{\int{\frac{1}{v} d v}}{2}\right)}}$$

The integral of $\frac{1}{v}$ is $\int{\frac{1}{v} d v} = \ln{\left(\left|{v}\right| \right)}$:

$$\frac{u^{6}}{6} - \frac{u^{4}}{4} + \frac{u^{2}}{2} - \frac{{\color{red}{\int{\frac{1}{v} d v}}}}{2} = \frac{u^{6}}{6} - \frac{u^{4}}{4} + \frac{u^{2}}{2} - \frac{{\color{red}{\ln{\left(\left|{v}\right| \right)}}}}{2}$$

Recall that $v=u^{2} + 1$:

$$\frac{u^{6}}{6} - \frac{u^{4}}{4} + \frac{u^{2}}{2} - \frac{\ln{\left(\left|{{\color{red}{v}}}\right| \right)}}{2} = \frac{u^{6}}{6} - \frac{u^{4}}{4} + \frac{u^{2}}{2} - \frac{\ln{\left(\left|{{\color{red}{\left(u^{2} + 1\right)}}}\right| \right)}}{2}$$

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

$$- \frac{\ln{\left(1 + {\color{red}{u}}^{2} \right)}}{2} + \frac{{\color{red}{u}}^{2}}{2} - \frac{{\color{red}{u}}^{4}}{4} + \frac{{\color{red}{u}}^{6}}{6} = - \frac{\ln{\left(1 + {\color{red}{\tan{\left(x \right)}}}^{2} \right)}}{2} + \frac{{\color{red}{\tan{\left(x \right)}}}^{2}}{2} - \frac{{\color{red}{\tan{\left(x \right)}}}^{4}}{4} + \frac{{\color{red}{\tan{\left(x \right)}}}^{6}}{6}$$

Therefore,

$$\int{\tan^{7}{\left(x \right)} d x} = - \frac{\ln{\left(\tan^{2}{\left(x \right)} + 1 \right)}}{2} + \frac{\tan^{6}{\left(x \right)}}{6} - \frac{\tan^{4}{\left(x \right)}}{4} + \frac{\tan^{2}{\left(x \right)}}{2}$$

Add the constant of integration:

$$\int{\tan^{7}{\left(x \right)} d x} = - \frac{\ln{\left(\tan^{2}{\left(x \right)} + 1 \right)}}{2} + \frac{\tan^{6}{\left(x \right)}}{6} - \frac{\tan^{4}{\left(x \right)}}{4} + \frac{\tan^{2}{\left(x \right)}}{2}+C$$

$\int \tan^{7}{\left(x \right)}\, dx = \left(- \frac{\ln\left(\tan^{2}{\left(x \right)} + 1\right)}{2} + \frac{\tan^{6}{\left(x \right)}}{6} - \frac{\tan^{4}{\left(x \right)}}{4} + \frac{\tan^{2}{\left(x \right)}}{2}\right) + C$A