Theresa Archer

2022-06-26

Integral of $\int {\mathrm{tan}}^{5}\left(x\right){\mathrm{sec}}^{4}\left(x\right)dx$?

Jaylee Dodson

Expert

Substituting $u=\mathrm{tan}\left(x\right),\phantom{\rule{thinmathspace}{0ex}}du={\mathrm{sec}}^{2}\left(x\right)\phantom{\rule{thinmathspace}{0ex}}dx$
$\begin{array}{rcl}\int {\mathrm{tan}}^{5}\left(x\right){\mathrm{sec}}^{4}\left(x\right)\phantom{\rule{thinmathspace}{0ex}}dx& =& \int {\mathrm{tan}}^{5}\left(x\right){\mathrm{sec}}^{2}\left(x\right){\mathrm{sec}}^{2}\left(x\right)\phantom{\rule{thinmathspace}{0ex}}dx\\ & =& \int {\mathrm{tan}}^{5}\left(x\right)\left[{\mathrm{tan}}^{2}\left(x\right)+1\right]{\mathrm{sec}}^{2}\left(x\right)\phantom{\rule{thinmathspace}{0ex}}dx\\ & =& \int {u}^{5}\left({u}^{2}+1\right)\phantom{\rule{thinmathspace}{0ex}}du\\ & =& \int {u}^{7}+{u}^{5}\phantom{\rule{thinmathspace}{0ex}}du\\ & =& \frac{1}{8}{u}^{8}+\frac{1}{6}{u}^{6}+c\\ & =& \frac{1}{8}{\mathrm{tan}}^{8}\left(x\right)+\frac{1}{6}{\mathrm{tan}}^{6}\left(x\right)+c\end{array}$

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