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I've completed the squares in order to get a fraction in the integrand of the form $\frac {1}{\sqrt{a^2-x^2}}$ that can be easily substituted by a trigonometric function (drawing the respective triangle...).

Doing the last step in the original function I got: $$\int \frac {1}{\sqrt {9-(x+2)^2}}dx$$ And substituting $\sqrt{9-(x+2)^2}=3\sin q$ and $dx=-3\sin q$, I got

$$\int \frac{-3\sin q}{(3\sin q)^5}dq= -\int \frac {1}{\sin^4(q)}dq = -\int \csc^4(q) .$$ So the question is how do I suppose to integrate $\csc^4 (q)$?

English is not my first language so I apologize for any mistakes.

share|cite|improve this question… – geraldgreen Nov 25 '11 at 1:00
There are a few typos, so I don't know what the problem is exactly. 1) If the problem in the title is right, then the completing the square is not; 2) In the first displayed line of the main post, the exponent is not right; 3) If the completed square was right I would make the substitution $x+2=3\sin q$; There are also constant problems if the exponent is $5/2$; 4) With the suggested substitution we do get an integral close to yours, with the more familiar $\sec$. A technique like the one posted in an answer works. – André Nicolas Nov 25 '11 at 1:38
@AndréNicolas. I think you are right. I should have checked to see if he completed the square properly before posting the answer. He only has to complete the square well and then he can use the answer I have posted. – smanoos Nov 25 '11 at 1:44
@smanoos: However, your do answer the question about how to integrate $\csc^4\theta$, and that will remain useful even if the numbers change. – André Nicolas Nov 25 '11 at 1:48

Using trigonometric identities, observe the following: $$\int \csc^4\theta d\theta=\int \csc^2\theta \csc^2\theta ~d\theta=\int(\csc^2\theta)(1+ \cot^2\theta) d\theta=\int\csc^2\theta d\theta+\int \csc^2\theta\cot^2\theta d\theta .$$
You probably know that $\int \csc^2\theta d\theta=-\cot\theta$. For the second integral, you can proceed as follows; Let $u=\cot\theta$, so that $du=-\csc^2\theta d\theta$. So you have $$\int \csc^2\theta\cot^2\theta d\theta=-\int u^2du=-\frac{u^3}{3}=-\frac{\cot^3\theta}{3} .$$ Remember to add a constant of integration.

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