Find the indefinite integral. \int \sqrt[3]{\tan x}\sec^{2}xdx

Find the indefinite integral.
$\int \sqrt[3]{\mathrm{tan}x}{\mathrm{sec}}^{2}xdx$

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Step 1
We have to solve the indefinite integral:
$\int \sqrt[3]{\mathrm{tan}x}{\mathrm{sec}}^{2}xdx$
Solving the integral by substitution method.
Assuming,
$t=\mathrm{tan}x$
Differentiating,
$\frac{dt}{dx}=\frac{d\mathrm{tan}x}{dx}$
$={\mathrm{sec}}^{2}x$
$dt={\mathrm{sec}}^{2}xdx$
Step 2
Substituting above values in the given integral, we get
$\int \sqrt[3]{\mathrm{tan}x}{\mathrm{sec}}^{2}xdx=\int \sqrt[3]{t}dt$
$=\int {t}^{\frac{1}{3}}dt$
$=\frac{{t}^{\frac{1}{3}+1}}{\frac{1}{3}+1}+C$ (since $\int {x}^{n}dx=\frac{{x}^{n+1}}{n+1}+C$)
$=\frac{{t}^{\frac{1+3}{3}}}{\frac{1+3}{3}}+C$
$=\frac{{t}^{\frac{4}{3}}}{\frac{4}{3}}+C$
$=\frac{3}{4}{t}^{\frac{4}{3}}+c$
$=\frac{3}{4}{\left(\mathrm{tan}x\right)}^{\frac{4}{3}}+C$
Hence, value of given indefinite integration is $\frac{3}{4}{\left(\mathrm{tan}x\right)}^{\frac{4}{3}}+C$.

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Elois Puryear
Given:
$\int {\mathrm{sec}}^{2}\left(x\right)\sqrt{3}\left\{\mathrm{tan}\left(x\right)\right\}dx$
Substitution $u=\mathrm{tan}\left(x\right)⇒\frac{du}{dx}={\mathrm{sec}}^{2}\left(x\right)$
$=\int \sqrt{3}\left\{u\right\}du$
Integral of a power function:
$\int {u}^{n}du=\frac{{u}^{n+1}}{n+1}$ at $n=\frac{1}{3}:$
$=\frac{3{u}^{\frac{4}{3}}}{4}$
Reverse replacement $u=\mathrm{tan}\left(x\right):$
$=\frac{3{\mathrm{tan}}^{\frac{4}{3}}\left(x\right)}{4}$
$\int {\mathrm{sec}}^{2}\left(x\right)\sqrt{3}\left\{\mathrm{tan}\left(x\right)\right\}dx$
$=\frac{3{\mathrm{tan}}^{\frac{4}{3}}\left(x\right)}{4}+C$