Match List-I with List-II (\(-2 \lt x \lt 2\), C is a constant of integration)
| List-I | List-II | ||
| (A) | \(\int\sqrt{4-x^2}\,dx\) | (I) | \(-\sqrt{4-x^2}+C\) |
| (B) | \(\int x\sqrt{4-x^2}\,dx\) | (II) | \(\displaystyle\sin^{-1}\frac{x}{2}+C\) |
| (C) | \(\int\displaystyle\frac{x}{\sqrt{4-x^2}}\,dx\) | (III) | \(\displaystyle\frac{x}{2}\sqrt{4-x^2}+2\sin^{-1}\frac{x}{2}+C\) |
| (D) | \(\int\displaystyle\frac{dx}{\sqrt{4-x^2}}\) | (IV) | \(\displaystyle -\frac{1}{3}\left(4-x^2\right)^{3/2}+C\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
All four contain \(\sqrt{4-x^2}\). When the factor x is present, it is half the derivative of \(4-x^2\), so a substitution works; without it, a standard integral with a = 2 is used.
(A) \(\int\sqrt{4-x^2}\,dx\)
This matches (III) \(\frac{x}{2}\sqrt{4-x^2}+2\sin^{-1}\frac{x}{2}+C\).
\( \displaystyle \int\sqrt{a^2-x^2}\,dx = \frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\frac{x}{a}+C \)
With a = 2, \(\frac{a^2}{2} = 2\).
(B) \(\int x\sqrt{4-x^2}\,dx\)
This matches (IV) \(-\frac{1}{3}\left(4-x^2\right)^{3/2}+C\).
Putting \(t = 4-x^2\), \(dt = -2x\,dx\):
\( \displaystyle \int x\sqrt{4-x^2}\,dx = -\frac{1}{2}\int t^{1/2}\,dt = -\frac{1}{3}t^{3/2}+C \)
(C) \(\int\frac{x}{\sqrt{4-x^2}}\,dx\)
This matches (I) \(-\sqrt{4-x^2}+C\).
With the same substitution:
\( \displaystyle \int\frac{x}{\sqrt{4-x^2}}\,dx = -\frac{1}{2}\int t^{-1/2}\,dt = -t^{1/2}+C \)
(D) \(\int\frac{dx}{\sqrt{4-x^2}}\)
This matches (II) \(\sin^{-1}\frac{x}{2}+C\).
\( \displaystyle \int\frac{dx}{\sqrt{a^2-x^2}} = \sin^{-1}\frac{x}{a}+C \)