# Evaluate the following integrals. inttan^3 4x dx

Evaluate the following integrals.
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$\text{To evaluate:}$

${\mathrm{tan}}^{2}x={\mathrm{sec}}^{2}x-1$
$\frac{d}{dx}\left(\mathrm{sec}x\right)=\mathrm{sec}x\mathrm{tan}x$
$\text{Consider}\int {\mathrm{tan}}^{3}4xdx$
$=\int {\mathrm{tan}}^{2}4x\mathrm{tan}4xdx$
$=\int \left({\mathrm{sec}}^{2}4x-1\right)\mathrm{tan}4xdx$

$⇒4\mathrm{sec}4x\mathrm{tan}4xdx=du$
$⇒\mathrm{tan}4xdx=\frac{du}{4\mathrm{sec}4x}$
$\text{Substituting all the values, we get,}$
$\int \left({u}^{2}-1\right)\frac{du}{\mathrm{sec}4x}$
$=\int \left({u}^{2}-1\right)\frac{du}{4u}$
$=\frac{1}{4}\int \frac{{u}^{2}-1}{u}du$
$=\frac{1}{4}\int u-\frac{1}{u}du$
$=\frac{1}{4}\left[\frac{{u}^{2}}{2}-\mathrm{ln}\left(u\right)\right]+C$
$=\frac{1}{4}\left[\frac{{\mathrm{sec}}^{2}4x}{2}-\mathrm{ln}\left(\mathrm{sec}4x\right)\right]+C$