Match the LIST-I with LIST-II
| LIST-I | LIST-II | ||
| Indefinite Integral | Solution (where c is an arbitrary constant) | ||
| A. | \( \int\sqrt{16-x^2}\,dx \) | I. | \( \sin^{-1}\left(\displaystyle\frac{x}{4}\right)+c \) |
| B. | \( \int\displaystyle\frac{dx}{\sqrt{16-x^2}} \) | II. | \( \displaystyle\frac{x}{2}\sqrt{16-x^2}+8\sin^{-1}\left(\frac{x}{4}\right)+c \) |
| C. | \( \int\displaystyle\frac{dx}{\sqrt{x^2-16}} \) | III. | \( \displaystyle\frac{1}{8}\log\left|\frac{4+x}{4-x}\right|+c \) |
| D. | \( \int\displaystyle\frac{dx}{16-x^2} \) | IV. | \( \log\left|x+\sqrt{x^2-16}\right|+c \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → A-II, B-I, C-IV, D-III
Each integral is a standard form with a = 4.
(A) \( \int\sqrt{16-x^2}\,dx \)
This matches (II).
\( \displaystyle \int\sqrt{a^2-x^2}\,dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\frac{x}{a}+c \)
\( \displaystyle =\frac{x}{2}\sqrt{16-x^2}+8\sin^{-1}\left(\frac{x}{4}\right)+c \)
(B) \( \int\frac{dx}{\sqrt{16-x^2}} \)
This matches (I).
\( \displaystyle \int\frac{dx}{\sqrt{a^2-x^2}}=\sin^{-1}\frac{x}{a}+c \)
\( \displaystyle =\sin^{-1}\left(\frac{x}{4}\right)+c \)
(C) \( \int\frac{dx}{\sqrt{x^2-16}} \)
This matches (IV).
\( \displaystyle \int\frac{dx}{\sqrt{x^2-a^2}}=\log\left|x+\sqrt{x^2-a^2}\right|+c \)
\( \displaystyle =\log\left|x+\sqrt{x^2-16}\right|+c \)
(D) \( \int\frac{dx}{16-x^2} \)
This matches (III).
\( \displaystyle \int\frac{dx}{a^2-x^2}=\frac{1}{2a}\log\left|\frac{a+x}{a-x}\right|+c \)
\( \displaystyle =\frac{1}{8}\log\left|\frac{4+x}{4-x}\right|+c \)
So the matching is A-II, B-I, C-IV, D-III.