Evaluate the improper integral.
-$$\int _ { 0 } ^ { 9 } \frac { d x } { \sqrt { 81 - x ^ { 2 } } }$$
A) 18
B) $\frac { \pi } { 18 }$
C) 1
D) $\frac { \pi } { 2 }$

Question

Quizplus · Accepted Answer

Let us use the substitution $x=9\sin(	heta)$, $dx=9\cos(	heta)d	heta$, and evaluate the integral over the interval $\left[0,\frac{\pi}{2}ight]$. Then, we have:

\begin{align*}
\int _ { 0 } ^ { 9 } \frac { d x } { \sqrt { 81 - x ^ { 2 } } } &= \int_{0}^{\frac{\pi}{2}} \frac{9\cos(	heta)}{\sqrt{81-81\sin^2(	heta)}}d	heta\
&=\int_{0}^{\frac{\pi}{2}} \frac{9\cos(	heta)}{9\cos(	heta)}d	heta\
&=\int_{0}^{\frac{\pi}{2}} d	heta\
&=\frac{\pi}{2}
\end{align*}

Thus, the correct answer is $\boxed{	extbf{(D)}\ \frac{\pi}{2}}$.

Evaluate the Improper Integral ∫09dx81−x2\int _ { 0 } ^ { 9 } \frac { d x } { \sqrt { 81 - x ^ { 2 } } }∫09​81−x2​dx​

Evaluate the Improper Integral $\int _ { 0 } ^ { 9 } \frac { d x } { \sqrt { 81 - x ^ { 2 } } }$