$\displaystyle\int \dfrac{1}{\sqrt{1 - x^2}}\,dx$ equals:
- $\log|1 - x^2| + C$
- $\sin^{-1} x + C$
- $\tan^{-1} x + C$
- $\cos^{-1} x + C$
Show answer and solution
Correct answer: B. $\sin^{-1} x + C$
$\int \dfrac{1}{\sqrt{a^2 - x^2}}\,dx = \sin^{-1}(x/a) + C$, which with $a = 1$ gives $\sin^{-1} x + C$.
The integral of the negative integrand would give $\cos^{-1} x + C$, since $\dfrac{d}{dx}\sin^{-1} x = +\dfrac{1}{\sqrt{1-x^2}}$ and $\dfrac{d}{dx}\cos^{-1} x = -\dfrac{1}{\sqrt{1-x^2}}$.