allery$\int \dfrac{dx}{\sqrt{a^2-x^2}}$ equals
A$\sin^{-1}\dfrac{x}{a}+C$
B$\dfrac{1}{a}\tan^{-1}\dfrac{x}{a}+C$
C$\log\left|x+\sqrt{a^2-x^2}\right|+C$
D$\cos^{-1}\dfrac{x}{a}+C$
Answer & Solution
Correct answer: A. $\sin^{-1}\dfrac{x}{a}+C$
1. Substitute $x=a\sin\theta$, so $dx=a\cos\theta\,d\theta$.
2. $\sqrt{a^2-x^2}=a\cos\theta$.
3. Integral $=\int d\theta=\theta=\sin^{-1}\dfrac{x}{a}+C$.
4. Check: $\dfrac{d}{dx}\sin^{-1}\dfrac{x}{a}=\dfrac{1}{\sqrt{a^2-x^2}}$.
_Source: syllabus Mathematics Ch 7 "Integrals", p.307_
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