To find an integral that satisfies a substitution property. Consider the following integral. $$\int \cfrac{1}{9-x^2}dx$$
Substituting $x=3\sin u$ makes the integral $\displaystyle\int\cfrac{1}{3\cos u}du$.
I would like to modify the original integral such that after the substitution $x=3\sin u$, it looks like $\displaystyle\int\cfrac{1}{9\cos^2u}du$.
My only thought so far was that $dx=3\cos u du$ which tells me that I need to have $\cos^3u$ in the denominator of the integrand to get $\cos^2u$ in the denominator after simplification. How I do proceed from here? thanks for your time.

Let $u\in[-\pi/2,\pi/2]$. After substitution in the original integral you get$$\int\frac{3\cos u~du}{9\cos^2u}$$so we want to get an additional $3\cos u$ term in the denominator to cancel the $3\cos u$ in the numerator. Note that $x=3\sin u\implies\sqrt{9-x^2}=3|\cos u|=3\cos u$, so modify the original integral to$$\int\frac{dx}{(9-x^2)\sqrt{9-x^2}}$$
