Holly Guerrero

2021-12-31

Evaluate the integral.
$\int {\mathrm{tan}}^{5}xdx$

Alex Sheppard

Step 1
Split ${\mathrm{tan}}^{5}x$ into simpler powers ${\mathrm{tan}}^{3}x$ and ${\mathrm{tan}}^{2}x$ and change ${\mathrm{tan}}^{2}x$ to ${\mathrm{sec}}^{2}x-1$ and simplify.
$\int {\mathrm{tan}}^{5}xdx=\int {\mathrm{tan}}^{3}x\left({\mathrm{tan}}^{2}xdx\right)$
$=\int {\mathrm{tan}}^{3}x\left({\mathrm{sec}}^{2}x-1\right)dx$
$=\int \left({\mathrm{tan}}^{3}x{\mathrm{sec}}^{2}x-{\mathrm{tan}}^{3}x\right)dx$
Step 2
use the integral theorem and integrate both the of $\int \left({\mathrm{tan}}^{3}x{\mathrm{sec}}^{2}x-{\mathrm{tan}}^{3}x\right)dx$ terms with respect to dx and simplify to obtain the value of the integral.
$\int \left({\mathrm{tan}}^{3}x{\mathrm{sec}}^{2}x-{\mathrm{tan}}^{3}x\right)dx=\int {\mathrm{tan}}^{3}x{\mathrm{sec}}^{2}xdx-\int {\mathrm{tan}}^{3}xdx$
$=\int \left({\mathrm{tan}}^{3}x{\mathrm{sec}}^{2}x\right)dx-\int {\mathrm{tan}}^{2}xdx\left(\mathrm{tan}x\right)dx$
$=\int {\mathrm{tan}}^{3}x{\mathrm{sec}}^{2}xdx-\int \left({\mathrm{sec}}^{2}x-1\right)\mathrm{tan}xdx$
$=\int {\mathrm{tan}}^{3}x{\mathrm{sec}}^{2}xdx-\int \mathrm{tan}x{\mathrm{sec}}^{2}xdx+\int \mathrm{tan}xdx$
$=\frac{1}{4}{\mathrm{tan}}^{4}x-\frac{1}{2}{\mathrm{tan}}^{2}x+\mathrm{ln}\mathrm{sec}x+C$

Linda Birchfield

$\int {\mathrm{tan}\left(x\right)}^{5}dx$
Evaluate the integral
$\frac{1}{4}×{\mathrm{tan}\left(x\right)}^{4}-\int {\mathrm{tan}\left(x\right)}^{3}dx$
Evaluate the integral
$\frac{1}{4}×{\mathrm{tan}\left(x\right)}^{4}-\left(\frac{1}{2}×{\mathrm{tan}\left(x\right)}^{2}-\int \mathrm{tan}\left(x\right)dx\right)$
Expand the expression
$\frac{1}{4}×{\mathrm{tan}\left(x\right)}^{4}-\left(\frac{1}{2}×{\mathrm{tan}\left(x\right)}^{2}-\int \frac{\mathrm{sin}\left(x\right)}{\mathrm{cos}\left(x\right)}dx\right)$
Transform the expression
$\frac{1}{4}×{\mathrm{tan}\left(x\right)}^{4}-\left(\frac{1}{2}×{\mathrm{tan}\left(x\right)}^{2}-\int -\frac{1}{t}dt\right)$
Use properties of integrals
$\frac{1}{4}×{\mathrm{tan}\left(x\right)}^{4}-\left(\frac{1}{2}×{\mathrm{tan}\left(x\right)}^{2}+\int \frac{1}{t}dt\right)$
Evaluate the integral
$\frac{1}{4}×{\mathrm{tan}\left(x\right)}^{4}-\left(\frac{1}{2}×{\mathrm{tan}\left(x\right)}^{2}+\mathrm{ln}\left(|t|\right)\right)$
Substitute back
$\frac{1}{4}×{\mathrm{tan}\left(x\right)}^{4}-\left(\frac{1}{2}×{\mathrm{tan}\left(x\right)}^{2}+\mathrm{ln}\left(|\mathrm{cos}\left(x\right)|\right)\right)$
Simplify
$\frac{{\mathrm{tan}\left(x\right)}^{4}}{4}-\frac{{\mathrm{tan}\left(x\right)}^{2}}{2}-\mathrm{ln}\left(|\mathrm{cos}\left(x\right)|\right)$
Solution
$\frac{{\mathrm{tan}\left(x\right)}^{4}}{4}-\frac{{\mathrm{tan}\left(x\right)}^{2}}{2}-\mathrm{ln}\left(|\mathrm{cos}\left(x\right)|\right)+C$

karton

$\begin{array}{}{\mathrm{tan}}^{5}\left(x\right)dx\\ =\int \left({\mathrm{tan}}^{2}\left(x\right){\right)}^{2}\mathrm{tan}\left(x\right)dx\\ =\int \left({\mathrm{sec}}^{2}\left(x\right)-1{\right)}^{2}\mathrm{tan}\left(x\right)dx\\ =\int \frac{\left({u}^{2}-1{\right)}^{2}}{u}du\\ =\int \left({u}^{3}-2u+\frac{1}{u}\right)du\\ =\int {u}^{3}du-2\int udu+\int \frac{1}{u}du\\ \int {u}^{3}du\\ =\frac{{u}^{4}}{4}\\ \int udu\\ =\frac{{u}^{2}}{2}\\ \int \frac{1}{u}du\\ =\mathrm{ln}\left(u\right)\\ \int {u}^{3}du-2\int udu+\int \frac{1}{u}du\\ =\mathrm{ln}\left(u\right)+\frac{{u}^{4}}{4}-{u}^{2}\\ =\mathrm{ln}\left(\mathrm{sec}\left(x\right)\right)+\frac{{\mathrm{sec}}^{4}\left(x\right)}{4}-{\mathrm{sec}}^{2}\left(x\right)\\ \int {\mathrm{tan}}^{5}\left(x\right)dx\\ =\mathrm{ln}\left(|\mathrm{sec}\left(x\right)|\right)+\frac{{\mathrm{sec}}^{4}\left(x\right)}{4}-{\mathrm{sec}}^{2}\left(x\right)+C\end{array}$