Integral by using substitution (How to proceed?)

Question

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?

Answer 1

Using your substition:

$$\int\frac1{x^2\sqrt{a^2-x^2}}\;\rightarrow\;\int\frac{a\cos\theta\;d\theta}{a^2\sin^2\theta\cdot a\cos\theta}=-\frac1{a^2}\int\frac{d\theta}{-\sin^2\theta}=-\frac1{a^2}\cot\theta+C$$

Answer 2

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: $$ I(a)=-\frac{\sqrt{a^2-x^2}}{ a^2 x}+C $$