<s>[INST] Problem in trigonometry integral $$\int \frac{1}{(9-x^2)^{\frac{3}{2}}}dx$$
Let $x=3\sin (u)$
$dx=3\cos (u)$
So, $(9-x^2)^{\frac{3}{2}}=(9-9\sin ^2(u))^{\frac{3}{2}}$
$=27\cos ^3(u)$
$u=\sin ^-1(\frac{x}{3})$
My problem is how to substitute u into the integral to become $3\int \frac{\sec ^2(u)}{27}$du?
 [/INST] You are pretty much there.
$\int \frac{1}{(9-x^2)^{\frac{3}{2}}}dx = \int \frac{3\cos u}{27 cos^3 u}du = \frac{1}{9}\int \sec^2 u \; du$
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