Example 6.3.2. Using Trigonometric Substitution.
Evaluate \(\ds \int_{-3}^3\sqrt{9-x^2}\, dx\text{.}\)
Solution 1.
We begin by noting that \(9\left(\sin^2(\theta) + \cos^2(\theta)\right) = 9\text{,}\) and hence \(9\cos^2(\theta) = 9-9\sin^2(\theta)\text{.}\) If we let \(x=3\sin(\theta)\text{,}\) then \(9-x^2 = 9-9\sin^2(\theta) = 9\cos^2(\theta)\text{.}\)
Setting \(x=3\sin(\theta)\) gives \(dx = 3\cos(\theta) \, d\theta\text{.}\) We are almost ready to substitute. We also wish to change our bounds of integration. The bound \(x=-3\) corresponds to \(\theta = -\pi/2\) (for when \(\theta = -\pi/2\text{,}\) \(x=3\sin(\theta) = -3\)). Likewise, the bound of \(x=3\) is replaced by the bound \(\theta = \pi/2\text{.}\) Thus
\begin{align*}
\int_{-3}^3\sqrt{9-x^2}\, dx \amp = \int_{-\pi/2}^{\pi/2} \sqrt{9-9\sin^2(\theta) }\,(3\cos(\theta) )\, d\theta\\
\amp = \int_{-\pi/2}^{\pi/2} 3\sqrt{9\cos^2(\theta) } \cos(\theta) \, d\theta\\
\amp =\int_{-\pi/2}^{\pi/2} 3\abs{3\cos(\theta) } \cos(\theta) \, d\theta\text{.}
\end{align*}
On \([-\pi/2,\pi/2]\text{,}\) \(\cos(\theta)\) is always positive, so we can drop the absolute value bars, then employ a power-reducing formula:
\begin{align*}
\int_{-3}^3\sqrt{9-x^2}\, dx \amp = \int_{-\pi/2}^{\pi/2} 9\cos^2(\theta) \, d\theta\\
\amp = \int_{-\pi/2}^{\pi/2} \frac{9}{2}\big(1+\cos(2\theta)\big)\, d\theta\\
\amp = \left.\frac92 \big(\theta +\frac12\sin(2\theta)\big)\right|_{-\pi/2}^{\pi/2}\\
\amp = \frac92\pi\text{.}
\end{align*}
This matches our answer from before.