Find the indefinite integral. $$\int \frac { 1 } { \sqrt { 9 - 16 x ^ { 2 } } } d x$$
A) $\frac { 1 } { 4 } \arcsin \left( \frac { 4 x } { 3 } \right) + C$
B) $\frac { 1 } { 4 } \arcsin ( 9 x ) + C$
C) $4 \arcsin ( 9 x ) + C$
D) $4 \arcsin \left( \frac { 4 x } { 3 } \right) + C$
E) $\frac { 1 } { 4 } \arcsin ( 16 x ) + C$

Question

Quizplus · Accepted Answer

This integral is of the form $\int \frac{1}{\sqrt{a^2 - u^2}} du$, which directly corresponds to the integral formula for $\arcsin(\frac{u}{a})$, with $a = 3$ and $u = 4x$, leading to the answer $\frac{1}{4} \arcsin(\frac{4x}{3}) + C$.

Find the Indefinite Integral ∫19−16x2dx\int \frac { 1 } { \sqrt { 9 - 16 x ^ { 2 } } } d x∫9−16x2​1​dx

Find the Indefinite Integral $\int \frac { 1 } { \sqrt { 9 - 16 x ^ { 2 } } } d x$