Find the integral using an appropriate trigonometric substitution. Select the correct answer.
$$\int \frac { x } { \sqrt { 1 - x ^ { 2 } } } d x$$
A) $\sqrt { 1 - x ^ { 2 } } + C$
B) $\sqrt { 1 - x } + C$
C) $- \sqrt { 1 - x ^ { 2 } } + C$
D) $- \sqrt { 1 - x } + C$

Question

Quizplus · Accepted Answer

We can make the substitution $x = \sin 	heta$. Then, $dx = \cos 	heta \, d	heta$ and $\sqrt{1-x^2} = \cos 	heta$. Substituting these into the integral, we get:
$$\int \frac{x}{\sqrt{1-x^2}} \, dx = \int \frac{\sin 	heta}{\cos 	heta} \cdot \cos 	heta \, d	heta = \int \sin 	heta \, d	heta = -\cos 	heta + C = -\sqrt{1-x^2} + C$$
Therefore, the correct choice is (C).

Find the Integral Using an Appropriate Trigonometric Substitution ∫x1−x2dx\int \frac { x } { \sqrt { 1 - x ^ { 2 } } } d x∫1−x2​x​dx

Find the Integral Using an Appropriate Trigonometric Substitution $\int \frac { x } { \sqrt { 1 - x ^ { 2 } } } d x$