Integration Problem with a Trig substitution

Question

Okay I am a little stuck on this problem.

$$\int \tan^5(x)\sqrt{\sec(x)} \; dx$$

What should be my first step for a u sub or a trig sub? I have tried to use $u=\sec(x)$ and then $u=\tan(x)$, but I get stuck. A little help?

Answer 1

Let $u=\sec x$. Then $du=\sec x\tan x\,dx$. Substitute. With part of the substitution done, we are looking at $$\int \frac{\tan^5 x}{\sec x\tan x}u^{1/2}\,du.$$ Using the fact that $\tan^2 x=\sec^2 x-1$, we end up with $$\int \frac{(u^2-1)^2}{u^{1/2}}\,du,$$ that is, $$\int (u^{7/2}-2u^{3/2}+u^{-1/2})\,du.$$

Answer 2

Let $u=\cos x$ then $du=\sin x dx$ and then the anti derivative becomes:

$$\int\frac{(1-u^2)^2}{u^5}\frac{du}{\sqrt u}$$ now let $u=t^2$ so we find $$2\int \frac{(1-t^4)^2}{t^{10}}dt=2\int t^{-10}dt-4\int t^{-6}dt+2\int t^{-2}dt$$ I'm sure that you can take it from here.