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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.

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    @smanoos: However, your do answer the question about how to integrate $\csc^4\theta$, and that will remain useful even if the numbers change.2011-11-25

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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.