<s>[INST] Integration Trig Substitution After making the correct trig substitution what does the integral of

$\dfrac{1}{\sqrt{9-x^2}} dx$ reduce to without solving the equation?

I reduced it down to the integral of $3\cos(\theta)d\theta$ is this correct?
 [/INST] Performing trig. substitution of $x=3 \sin \theta$ (so $dx=3\cos \theta \, d\theta$) gives $$\int\frac{1}{\sqrt{9-x^2}} \, dx=\int\frac{1}{\sqrt{9-9 \sin^2 \theta}} \, (3 \cos \theta \, d\theta)=\int\frac{1}{|3 \cos \theta|} \, (3 \cos \theta \, d\theta) = \int 1 \, d\theta$$
which is equal to $\theta + C$, or $\displaystyle\sin^{-1} \left( \frac{x}{3} \right) + C$ after back-substitution.
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