Question

Can someone help me evaluate the integral tan^3xsec^3xdx

Answer 1

Reduce the power and recall the Pythagorean identity:
\[\begin{align*}\tan^3x\sec^3x&=\tan^2x\sec^2x\sec x\tan x\\
&=(\sec^2x-1)\sec^2x\sec x\tan x
\end{align*}\]Substitute \(u=\sec x\), now you have
\[\int\tan^3x\sec^3x\,dx=\int (u^2-1)u^2\,du\]