allery$\displaystyle\int \dfrac{1}{\sqrt{1 - x^2}}\,dx$ equals:
A$\log|1 - x^2| + C$
B$\sin^{-1} x + C$
C$\tan^{-1} x + C$
D$\cos^{-1} x + C$
Answer & 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 $\dfrac{d}{dx}\cos^{-1} x = -\dfrac{1}{\sqrt{1-x^2}}$.
Keep practising Unit 6A: Indefinite Integrals & Fundamental Theorem
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