Integration by substitution starting from a result

Question

I know: $$\int\frac{1}{x^2\sqrt{a^2-x^2}}=-\frac{\sqrt{a^2-x^2}}{a^2x}$$

and I want to calculate:

$$\int\frac{1}{x^2\sqrt{3-16x^2}}$$

I think I have to apply a substitution but i don't know how to deal with the $16$.

Answer 1

$$\int\frac{1}{x^2\sqrt{3-16x^2}}=\int\frac{1}{x^2\sqrt{16(3/16-x^2)}}$$ $$=\int\frac{1}{4x^2\sqrt{3/16-x^2}}=\frac{1}{4}\int\frac{1}{x^2\sqrt{3/16-x^2}}$$

Set $a^2=3/16$ and apply your formula.

Answer 2

You could let $\color{blue}{u=4x}$, so $du=4\,dx$ and $x=\frac{u}{4}$ to get

$\displaystyle\int\frac{1}{x^2\sqrt{3-16x^2}}\,dx=\frac{1}{4}\int\frac{1}{\left(u^2/16\right)\sqrt{3-u^2}}\,du=4\int\frac{1}{u^2\sqrt{3-u^2}}\,du$,

and then use your formula with $a=\sqrt{3}$.