Evaluating the definite integral $\int_0^3 \sqrt{9- x^2} \, dx$ I have been having a problem with the following definite integral:
$$\int_0^3 \sqrt{9- x^2} \, dx
$$
I am only familiar with u-substitution and am positive that it can be done with only that. Any help would be appreciated.

Here's a quick method: If $y = \sqrt{9-x^2}$ then $y^2 = 9-x^2$ so $x^2+y^2=9$.  If you know that $x^2 + y^2 = 9$ is the equation of a circle of radius $3$ centered at $(0,0)$, that means $y^2=9-x^2$ is also an equation of that circle, as is $y = \pm\sqrt{9-x^2}$.  So $y = \sqrt{9-x^2}$ (without $\text{“}\pm\text{''}$) is the top half of the circle.  So the integral is just the area of one quarter of the circle.  It you know the area is $\pi r^2$ and $r=3$, that tells you the answer.
(Of course, this won't work if the point is to prove that $\pi r^2$ is the area.)
Another method:
\begin{align}
x & = 3\sin\theta \\
dx & = 3\cos\theta\,d\theta \\
\sqrt{9-x^2} & = \sqrt{9 - 9\sin^2\theta} = \sqrt{9} \sqrt{1-\sin^2\theta} = 3\cos\theta \\[6pt]
& \text{As $x$ goes from $0$ to $3$ then $\sin\theta$ goes from $0$ to $1$,} \\
& \text{so $\theta$ goes from $0$ to $\pi/2$.}
\end{align}
Hence the integral becomes
\begin{align}
& 9\int_0^{\pi/2} \cos^2\theta\,d\theta = 9\int_{\pi/2}^0 \cos^2\left(\frac\pi2 - \zeta\right) \,(-d\zeta) & & (\text{The substitution is $\theta=\frac\pi2-\zeta$}) \\[10pt]
= {} & 9 \int_{\pi/2}^0 (\sin^2\zeta) \,(-d\zeta) & & \Big(\text{since $\cos\left(\frac\pi2 - \zeta\right)= \sin\zeta$}\Big) \\[10pt]
= {} & 9 \int_0^{\pi/2} (\sin^2\zeta)\,d\zeta \\[10pt]
= {} & 9 \int_0^{\pi/2} \sin^2\theta\,d\theta.
\end{align}
So we have $$ 9 \int_0^{\pi/2} \cos^2\theta\,d\theta = 9 \int_0^{\pi/2} \sin^2\theta\,d\theta. $$
But on the other hand
$$
9 \int_0^{\pi/2} \cos^2\theta\,d\theta + 9 \int_0^{\pi/2} \sin^2\theta\,d\theta = 9 \int_0^{\pi/2} 1\,d\theta \qquad\qquad(\text{since } \cos^2\theta+\sin^2\theta =1.)
$$
The latter integral comes to $9\pi/2$, so each of the two integrals separately must be half of that, or $9\pi/4$.
