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?

Using your substitutions:
$$
\int \frac{dx}{(9 - x^2)^{\frac{3}{2}}} \;\; =\;\; \int \frac{3 \cos u du}{27(1- \sin^2u)^{\frac{3}{2}}} \;\; =\;\; \frac{1}{9} \int\frac{\cos u}{\cos^3u} du \;\; =\;\; \frac{1}{9} \int \frac{du}{\cos^2u}.
$$
Now use the fact that $\sec u = \frac{1}{\cos u}$.
