# Calc1.64

$\int \tan ^{2}x\sec ^{4}x\,dx\,$

The procedure with $\tan x$ and $\sec x$ is pretty much the same as with $\sin x$ and $\cos x$. In both cases, we want to get to a place where we can do a substitution. With $\sin x$, a suitable $du$ is $\cos x$. This is the difference.

${\frac {d(\tan x)}{dx}}=\sec ^{2}x\,$

${\frac {d(\sec x)}{dx}}=\sec x\tan x\,$

So, if you think about it, you can figure out the procedure here, or for any other problem of this type, yourself. When there is an even number of $\sec x$, turn all but 2 into $\tan x$ and we'll have a whole bunch of $\tan x$ and the derivative of $\tan x$ and nothing else.

$\int \tan ^{2}x\sec ^{4}x\,dx=\int \tan ^{2}x(\tan ^{2}x+1)\sec ^{2}x\,dx=\int (\tan ^{4}x+\tan ^{2}x)\sec ^{2}x\,dx\,$

If we were to do a substitution, we'd do $u=\tan x$ and $du=\sec ^{2}xdx$ but we've done enough to not write out the steps.

$\int (\tan ^{4}x+\tan ^{2}x)\sec ^{2}x\,dx={\frac {1}{5}}\tan ^{5}x+{\frac {1}{3}}\tan ^{3}x+C\,$