$\text { Evaluate the integral } \int _ { 0 } ^ { 1 / 6 } \frac { 16 } { \sqrt { 1 - 9 x ^ { 2 } } } d x$
A) $\frac { 16 } { 25 \pi }$
B) $\frac { 8 } { 9 } \pi$
C) $\frac { 4 } { 9 \pi }$
D) $\frac { 16 } { 25 } \pi$
E) $\frac { 8 } { 9 \pi }$

Question

$\text { Evaluate the integral } \int _ { 0 } ^ { 1 / 6 } \frac { 16 } { \sqrt { 1 - 9 x ^ { 2 } } } d x$ 
A) $\frac { 16 } { 25 \pi }$
B) $\frac { 8 } { 9 } \pi$
C) $\frac { 4 } { 9 \pi }$
D) $\frac { 16 } { 25 } \pi$
E) $\frac { 8 } { 9 \pi }$

Quizplus · Accepted Answer

The integral is of the form $\int \frac{a}{\sqrt{1-(bx)^2}} dx$, which is solved using the arcsin function. Here, $a = 16$ and $b = 3$. The antiderivative is $\frac{a}{b} \arcsin(bx)$. Evaluating from 0 to $\frac{1}{6}$, we get $\frac{16}{3} \arcsin(\frac{1}{2}) = \frac{16}{3} \cdot \frac{\pi}{6} = \frac{8}{9} \pi$.

Evaluate the integral ∫01/6161−9x2dx\text { Evaluate the integral } \int _ { 0 } ^ { 1 / 6 } \frac { 16 } { \sqrt { 1 - 9 x ^ { 2 } } } d x Evaluate the integral ∫01/6​1−9x2​16​dx

$\text { Evaluate the integral } \int _ { 0 } ^ { 1 / 6 } \frac { 16 } { \sqrt { 1 - 9 x ^ { 2 } } } d x$