$\int \sqrt{a^2-x^2}\,dx$ equals
A$\dfrac{x}{2}\sqrt{a^2-x^2}+\dfrac{a^2}{2}\sin^{-1}\dfrac{x}{a}+C$
B$\dfrac{x}{2}\sqrt{a^2-x^2}-\dfrac{a^2}{2}\sin^{-1}\dfrac{x}{a}+C$
C$\dfrac{x}{2}\sqrt{a^2-x^2}+\dfrac{a^2}{2}\log\left|x+\sqrt{a^2-x^2}\right|+C$
D$-\dfrac{1}{3}(a^2-x^2)^{3/2}+C$
Answer & Solution
Correct answer: A. $\dfrac{x}{2}\sqrt{a^2-x^2}+\dfrac{a^2}{2}\sin^{-1}\dfrac{x}{a}+C$
1. Integrating by parts with 1 as second function gives the standard result.
2. For $\sqrt{a^2-x^2}$ the inverse-sine term appears, not a log.
3. $\int \sqrt{a^2-x^2}\,dx=\dfrac{x}{2}\sqrt{a^2-x^2}+\dfrac{a^2}{2}\sin^{-1}\dfrac{x}{a}+C$.
4. Option C wrongly uses a log term (that belongs to $\sqrt{x^2\pm a^2}$).
_Source: NCERT Class 12 Mathematics Ch 7 "Integrals", p.330_
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