Concepcion Hale

2022-01-07

Evaluate the integral.
$\mathrm{tan}x{\mathrm{sec}}^{3}xdx$

Alex Sheppard

Step 1
Given
Evaluate the integral.
$\int \mathrm{tan}x{\mathrm{sec}}^{3}xdx$
Step 2
Let $u=\mathrm{sec}\left(x\right).$
Then
$du={\left(\mathrm{sec}\left(x\right)\right)}^{\prime }dx$
$=\mathrm{tan}\left(x\right)\mathrm{sec}\left(x\right)dxdu$
$={\left(\mathrm{sec}\left(x\right)\right)}^{\prime }dx$
$=\mathrm{tan}\left(x\right)\mathrm{sec}\left(x\right)dx$
,and we have that
$\mathrm{tan}\left(x\right)\mathrm{sec}\left(x\right)dx=du.$
Therefore,
$\int \mathrm{tan}\left(x\right){\mathrm{sec}}^{3}\left(x\right)dx=\int {u}^{2}du$
Apply the power rule $\int {u}^{n}du=\frac{{u}^{n+1}}{n+1}\left(n\ne -1\right)$
$\int {u}^{2}du=\frac{{u}^{1+2}}{1+2}=\left(\frac{{u}^{3}}{3}\right)$
Recall that $u=\mathrm{sec}\left(x\right)u=\mathrm{sec}\left(x\right):$
$\frac{{u}^{3}}{3}=\frac{{\mathrm{sec}\left(x\right)}^{3}}{3}$
Therefore,
$\int \mathrm{tan}\left(x\right){\mathrm{sec}}^{3}\left(x\right)dx=\frac{{\mathrm{sec}}^{3}\left(x\right)}{3}$
$\int \mathrm{tan}\left(x\right){\mathrm{sec}}^{3}\left(x\right)dx=\frac{{\mathrm{sec}}^{3}\left(x\right)}{3}+C$

$\int \mathrm{tan}x{\mathrm{sec}}^{3}xdx=\int {\mathrm{sec}}^{2}x\mathrm{tan}x\mathrm{sec}xdx=\left[\begin{array}{c}u=\mathrm{sec}x\\ du=\mathrm{tan}x\mathrm{sec}xdx\end{array}\right]$
$=\int {u}^{2}du=\frac{{u}^{3}}{3}+C=\frac{{\mathrm{sec}}^{3}x}{3}+C$
Result:
$\frac{{\mathrm{sec}}^{3}x}{3}+C$

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