The integral $\int \frac { 1 } { \sqrt { 4 x - x ^ { 2 } } } d x$ is

A) $\frac { \sin ^ { - 1 } \left( \frac { x - 2 } { 2 } \right) } { 4 } + C$
B) $- \sin ^ { - 1 } \left( \frac { x - 2 } { 2 } \right) + C$
C) $\frac { \sin ^ { - 1 } \left( \frac { x - 2 } { 2 } \right) } { 2 } + C$
D) $\sin ^ { - 1 } \left( \frac { x - 2 } { 2 } \right) + C$
E) $- \frac { \sin ^ { - 1 } \left( \frac { x - 2 } { 2 } \right) } { 2 } + C$

Question

The integral $\int \frac { 1 } { \sqrt { 4 x - x ^ { 2 } } } d x$ is

A) $\frac { \sin ^ { - 1 } \left( \frac { x - 2 } { 2 } \right) } { 4 } + C$
B) $- \sin ^ { - 1 } \left( \frac { x - 2 } { 2 } \right) + C$
C) $\frac { \sin ^ { - 1 } \left( \frac { x - 2 } { 2 } \right) } { 2 } + C$
D) $\sin ^ { - 1 } \left( \frac { x - 2 } { 2 } \right) + C$
E) $- \frac { \sin ^ { - 1 } \left( \frac { x - 2 } { 2 } \right) } { 2 } + C$

Quizplus · Accepted Answer

The integral can be solved by completing the square in the denominator: $4x - x^2 = -(x^2 - 4x) = -(x-2)^2 + 4$. This transforms the integral into a standard form $\int \frac{1}{\sqrt{a^2 - u^2}} du$, which results in $\sin^{-1}(\frac{u}{a}) + C$. Here, $u = x - 2$ and $a = 2$, leading to the solution $\sin^{-1}(\frac{x-2}{2}) + C$.

The Integral ∫14x−x2dx\int \frac { 1 } { \sqrt { 4 x - x ^ { 2 } } } d x∫4x−x2​1​dx

The Integral $\int \frac { 1 } { \sqrt { 4 x - x ^ { 2 } } } d x$