Question

Integration problem

Answer 1

\[\LARGE \int tan^5 x~ dx\]

Answer 2

I change \(\tan^5 x\) into \((tan^2x)^2~tanx\) right?

Answer 3

There's a reduction formula for this form I think:
\[\tan^5x=\tan^2x\tan^3x=(\sec^2x-1)\tan^3x\]
Do something similar with the cubed term. In the end, you'll be substituting \(u=\tan x\) so that \(du=\sec^2x~dx\).

Answer 4

\[ \tan^5 x=\tan^3 x (\sec^2x-1)=\tan^3x \sec^2x-\tan^3x\]\[\tan^3 x = \tan x (\sec^2x -1)=\tan x \sec^2x -\tan x\] \[\int\limits \tan^3x \sec^2x -\tan x \sec^2x + \tan x \ dx\ \]

for the left two terms take tanx=u, for the plain tanx, just turn it into sinx/cosx and pick v=cosx.

Answer 5

Thank you both :)