<s>[INST] Integral by using substitution (How to proceed?) Using the substitution $x=a\sin\theta$, or otherwise, find $\int\frac{1}{x^2\sqrt{a^2-x^2}}dx$.
My attempt, 
$x=a\sin\theta$
$dx=a\cos (\theta)d\theta$. Then $\sqrt{a^2-x^2}=\sqrt{a^2-a^2\sin ^2(\theta)}$
The given answer is $-\frac{\sqrt{a^2-x^2}}{a^2x}+c$ 
How to proceed then?
 [/INST] Other possibility:
Use  $x=1/y$, the integral becomes:
$$
I(a)=\int -\frac{1}{y^2}\frac{1}{\frac{1}{y^2} \sqrt{a^2-\frac{1}{y^2}}}dy=-\int \frac{y}{\sqrt{a^2y^2-1}}dy=-\frac{1}{a^2}\int \partial_y\left(\sqrt{a^2y^2-1}\right)dy=-\frac{\sqrt{a^2y^2-1}}{a^2}+C
$$
Resubstitute:
$$
I(a)=-\frac{\sqrt{a^2-x^2}}{ a^2 x}+C
$$
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