evaluate the following integral.\n \frac{3sqrt{3}}{2}int_{0}\frac{dx}{sqrt{9 - x^{2}}}\nwhich substitution…

Explanation:

Step1: Choose substitution

We use the substitution $x = 3\sin\theta$ because if $x = 3\sin\theta$, then $dx=3\cos\theta d\theta$ and $\sqrt{9 - x^{2}}=\sqrt{9 - 9\sin^{2}\theta}=3\cos\theta$. Also, when $x = 0$, $0 = 3\sin\theta$ gives $\theta=0$; when $x=\frac{3\sqrt{3}}{2}$, $\frac{3\sqrt{3}}{2}=3\sin\theta$ gives $\sin\theta=\frac{\sqrt{3}}{2}$ and $\theta=\frac{\pi}{3}$.

Step2: Rewrite the integral

The integral $\int_{0}^{\frac{3\sqrt{3}}{2}}\frac{dx}{\sqrt{9 - x^{2}}}$ becomes $\int_{0}^{\frac{\pi}{3}}\frac{3\cos\theta d\theta}{3\cos\theta}$.

Step3: Simplify the integral

$\int_{0}^{\frac{\pi}{3}}\frac{3\cos\theta d\theta}{3\cos\theta}=\int_{0}^{\frac{\pi}{3}}d\theta$.

Step4: Evaluate the integral

$\int_{0}^{\frac{\pi}{3}}d\theta=\left[\theta\right]_{0}^{\frac{\pi}{3}}=\frac{\pi}{3}-0=\frac{\pi}{3}$.

Answer:

$\frac{\pi}{3}$

Multiple - choice answer:

C. $x = 3\sin\theta$