<s>[INST] Evaluate $\int\frac{\sqrt {25 - x^2}}{ x^4}$ I'm pretty sure the method used is trig substitution. But I'm having trouble setting up and solving the problem. 
 [/INST] Subsititute $\theta=\sin^{-1} \dfrac{x}5\implies \dfrac{d\theta}{dx}=\dfrac{1}{\sqrt{25-x^2}}$, so we get,
\begin{align}
\\\\\\&\int\frac{\sqrt{25-x^2}}{x^4}dx
\\=&\int\frac{25-x^2}{x^4}\dfrac{1}{\sqrt{25-x^2}}d\theta
\\=&\int \dfrac{25-25\sin^2 \theta}{625\sin^4\theta}d\theta
\\=&\dfrac{1}{25}\int\dfrac{\cos^2 \theta}{\sin^4\theta}d\theta
\\=&\dfrac{1}{25}\int\cot^2\theta \csc^2\theta d\theta
\end{align}
Now subsitute $u=\cot \theta$ to finish.
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