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

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

Question

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

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

Quizplus · Accepted Answer

This integral can be solved by using trigonometric substitution. Let $x = 2 + 2\sin(\theta)$, which simplifies the square root to $2\cos(\theta)$ and dx to $2\cos(\theta)d\theta$. After substitution and simplification, the integral becomes a simple trigonometric integral, leading to the solution $- \sqrt { 4 x - x ^ { 2 } } + 2 \sin ^ { - 1 } \left( \frac { x - 2 } { 2 } \right) + C$.

The Integral ∫x4x−x2dx\int \frac { x } { \sqrt { 4 x - x ^ { 2 } } } d x∫4x−x2​x​dx

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