# Evaluate the integral. \int\sec^3x\tan^4x dx

Evaluate the integral.
$\int {\mathrm{sec}}^{3}x{\mathrm{tan}}^{4}xdx$
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tabuordg
The given integral is:
$I=\int {\mathrm{sec}}^{3}x{\mathrm{tan}}^{4}xdx$
$=\int {\mathrm{sec}}^{3}x{\left({\mathrm{tan}}^{2}x\right)}^{2}dx$
$=\int {\mathrm{sec}}^{3}x{\left({\mathrm{sec}}^{2}x-1\right)}^{2}dx$
$=\int {\mathrm{sec}}^{3}x\left({\mathrm{sec}}^{4}x-2{\mathrm{sec}}^{2}x+1\right)dx$
$=\int {\mathrm{sec}}^{7}xdx-2\int {\mathrm{sec}}^{5}xdx+\int {\mathrm{sec}}^{3}xdx$
Solving each of the component integrals we get,
$I=\frac{\mathrm{ln}\left(\mathrm{tan}x+\mathrm{sec}x\right)}{16}+\frac{{\mathrm{sec}}^{5}x\mathrm{tan}x}{6}-\frac{7{\mathrm{sec}}^{3}x\mathrm{tan}x}{24}+\frac{\mathrm{sec}x\mathrm{tan}x}{16}+C$
where C is the constant of integration.