# What's the integration of $\int \tan^6(x)\sec^4(x) dx$

I know I need to use $1+\tan^2(x) = \sec^2(x)$, but I don't know what to do next ? How do I get my $U$ variable so I can use $U$ substitution?

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You know the derivative of $\tan x$ is $\sec^2 x$, so if you can write it as some power of $\tan x$ times $\sec^2 x$ (or a couple of those added together), you are good to go. –  Graphth Feb 3 '13 at 21:25

I would let $U:=\tan(x)$, say. Then, as the comment says, $dU=\sec^2(x)dx$.

So, we have $\int\tan^6(x)\sec^4(x)dx = \int U^6(1+U^2)dU$.

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Since the derivative of $\tan x$ is $\sec^2x$, you want to have all tangents except for one factor of $\sec^2x$ that will be part of your $du$. Use the identity that you mentioned to convert two of the secants to tangents:

$$\tan^6x\sec^4x=\tan^6x\sec^2x\sec^2x=\tan^6x\left(1+\tan^2x\right)\sec^2x\;.$$

Now let $u=\tan x$, $du=\sec^2x~dx$, and your integral becomes

$$\int u^6\left(1+u^2\right)du\;.$$

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