Help solving $\int_{0}^{\frac{\pi}{6}} \frac{1}{\cos(u)^9}du$

Question

My calculus final is coming up. My teacher was nice enough to give old exams as study material, which I have been doing.

One questions asks: $$\int_{0}^{\frac{1}{2}} \frac{1}{(1-x^2)^5}dx$$ Using trigonometric substitution, I have found: $$\int_{0}^{\frac{\pi}{6}} \frac{1}{\cos(u)^9}du$$ However, I am not able to move forward. None of the integration techniques we have learnt seem to apply here. I have tried some online integral calculators which all show very complex approaches which we have not learnt. Assuming that was just how the algorithm solved it and that a human would have a more elegant solution, I came here for help.

Edit:

Turns out there is a typo in the question. It should read: $$\int_{0}^{\frac{1}{2}} \frac{1}{(1-x^2)^\frac{5}{2}}dx$$ That is much easier to solve.

I am going to leave this here just in case someone can answer it because that answer might be useful to someone at some point.

Answer 1

If you or your teacher insist on using trig method, then it will be long to finish it. The start would be using $1 = \left(\sin^2 \theta + \cos^2 \theta\right)^5$, and try to expand this binomially then simplify with the denominator $\cos^9 \theta$. It should work.

Answer 2

Hint:

I think you could try a hyperbolic substitution here (it's some form of trigonometry, after all): set $\;x=\tanh u, \enspace\mathrm d\mkern1mu x=(1-\tanh^2u)\,\mathrm d\mkern1mu u=\dfrac1{\cosh^2u}\,\mathrm d\mkern1mu u $. You should get $$\int_{0}^{\frac{1}{2}} \frac{1}{(1-x^2)^5}\,\mathrm d\mkern1mu x=\int_{0}^{\frac{\ln 3}{2}}\frac{\mathrm d\mkern1mu u}{(1-\tanh^2u)^4}=\int_{0}^{\frac{\ln 3}{2}}\cosh^8u\,\mathrm d\mkern1mu u.$$ There remains to linearise $\cosh^8u$, or simply express it with $\mathrm e^u$.