Question

what is the indefinite integral of tan^5xdx?

Answer 1

$\displaystyle\int\tan^5x\,\mathrm dx=\int\tan^4x\tan x$
$=\displaystyle\int(\sec^2x-1)^2\tan x\,\mathrm dx$
$=\displaystyle\int(\sec^4x-2\sec^2x+1)\tan x\,\mathrm dx$
$=\displaystyle\int\sec^3x\sec x\tan x\,\mathrm dx-2\int\sec^2x\tan x\,\mathrm dx+\int\tan x\,\mathrm dx$

For the first integral, let $a=\sec x$ so that $\mathrm da=\sec x\tan x\,\mathrm dx$. For the second, $b=\tan x$ so that $\mathrm db=\sec^2x\,\mathrm dx$. The last integral is standard, but if you're not familiar with it, you could write $\tan x=\dfrac{\sin x}{\cos x}$ and set $c=\cos x$ so that $\mathrm dc=-\sin x\,\mathrm dx$. Now

$=\displaystyle\int a^3\,\mathrm da-2\int b\,\mathrm db-\int\frac{\mathrm dc}c$
$=\displaystyle\frac{a^4}4-b^2-\ln|c|+C$
$=\displaystyle\frac{\sec^4x}4-\tan^2x-\ln|\cos x|+C$