# Evaluate the integral. \int x\sec x\tan x dx

Evaluate the integral.
$\int x\mathrm{sec}x\mathrm{tan}xdx$
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Step 1
u=x
du=dx
$dv=\mathrm{sec}x\cdot \mathrm{tan}x\cdot dx$
$v=\int \mathrm{sec}x\cdot \mathrm{tan}x\cdot dx$
$v=\mathrm{sec}x$
Step 2
Then we integrate by parts
$\int x\cdot \mathrm{sec}x\cdot \mathrm{tan}x\cdot dx$
$=\int u\cdot dv$
$=uv-\int v\cdot du$
$=x\cdot \mathrm{sec}x-\int \mathrm{sec}xdx$
$=x\cdot \mathrm{sec}x-\mathrm{ln}|\mathrm{sec}x+\mathrm{tan}x|+C$
Answer: $x\cdot \mathrm{sec}x-\mathrm{ln}|\mathrm{sec}x+\mathrm{tan}x|+C$
C=integrating constant