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

Explanation:

Step1: Recall the integral formula

The integral $\int\frac{dx}{\sqrt{a^{2}-x^{2}}}=\arcsin(\frac{x}{a})+C$, where $a > 0$. Here $a = 3$.

Step2: Apply the fundamental theorem of calculus

$\int_{0}^{\frac{3\sqrt{3}}{2}}\frac{dx}{\sqrt{9 - x^{2}}}=\left[\arcsin(\frac{x}{3})\right]_{0}^{\frac{3\sqrt{3}}{2}}$.

Step3: Evaluate the definite - integral

$\arcsin(\frac{\frac{3\sqrt{3}}{2}}{3})-\arcsin(0)$. Since $\frac{\frac{3\sqrt{3}}{2}}{3}=\frac{\sqrt{3}}{2}$ and $\arcsin(\frac{\sqrt{3}}{2})=\frac{\pi}{3}$, $\arcsin(0) = 0$. So the result is $\frac{\pi}{3}-0=\frac{\pi}{3}$.

Answer:

$\frac{\pi}{3}$