miklintisyt

2022-10-25

Solve:
${\int }_{0}^{1/3}{\mathrm{sec}}^{3}\left(\pi x\right)\phantom{\rule{thinmathspace}{0ex}}dx$

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Jovanni Salinas

Expert

It can be shown that
$\int {\mathrm{sec}}^{3}\theta d\theta =\frac{\mathrm{sec}\theta \mathrm{tan}\theta }{2}+\frac{\mathrm{ln}|\mathrm{sec}\theta +\mathrm{tan}\theta |}{2}+c.$
Going back to your integral, we let $\theta =\pi x$ and we get
$F\left(x\right)=\int {\mathrm{sec}}^{3}\left(\pi x\right)\phantom{\rule{thinmathspace}{0ex}}dx=\frac{\mathrm{sec}\left(\pi x\right)\mathrm{tan}\left(\pi x\right)}{2\pi }+\frac{\mathrm{ln}|\mathrm{sec}\left(\pi x\right)+\mathrm{tan}\left(\pi x\right)|}{2\pi }+c.$
Thus,
${\int }_{0}^{1/3}{\mathrm{sec}}^{3}\left(\pi x\right)\phantom{\rule{thinmathspace}{0ex}}dx=F\left(1/3\right)-F\left(0\right)$
and we are done.

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