The antiderivative of $f ( x ) = \frac { 2 } { \sqrt { 1 - x ^ { 2 } } } - \sin x$ is

A) $2 \sin ^ { - 1 } x - \cos x + C$
B) $2 \cos ^ { - 1 } x - \cos x + C$
C) $2 \sin ^ { - 1 } x + \cos x + C$
D) $- 2 \sin ^ { - 1 } x + \cos x + C$
E) $- 2 \sin ^ { - 1 } x + \sin x + C$

Question

The antiderivative of $f ( x ) = \frac { 2 } { \sqrt { 1 - x ^ { 2 } } } - \sin x$ is

A) $2 \sin ^ { - 1 } x - \cos x + C$
B) $2 \cos ^ { - 1 } x - \cos x + C$
C) $2 \sin ^ { - 1 } x + \cos x + C$
D) $- 2 \sin ^ { - 1 } x + \cos x + C$
E) $- 2 \sin ^ { - 1 } x + \sin x + C$

Quizplus · Accepted Answer

The Answer of The antiderivative of $f ( x ) = \frac { 2 } { \sqrt { 1 - x ^ { 2 } } } - \sin x$ is

A) $2 \sin ^ { - 1 } x - \cos x + C$
B) $2 \cos ^ { - 1 } x - \cos x + C$
C) $2 \sin ^ { - 1 } x + \cos x + C$
D) $- 2 \sin ^ { - 1 } x + \cos x + C$
E) $- 2 \sin ^ { - 1 } x + \sin x + C$

The Antiderivative Of f(x)=21−x2−sin⁡xf ( x ) = \frac { 2 } { \sqrt { 1 - x ^ { 2 } } } - \sin xf(x)=1−x2​2​−sinx

The Antiderivative Of $f ( x ) = \frac { 2 } { \sqrt { 1 - x ^ { 2 } } } - \sin x$