Question

Match the LIST-I with LIST-II

LIST-I Indefinite Integral	LIST-II Solution (where \(c\) is an arbitrary constant)
A. \(\int\sqrt{16-x^2}\,dx\)	I. \(\sin^{-1}\left(\frac{x}{4}\right)+c\)
B. \(\int\frac{dx}{\sqrt{16-x^2}}\)	II. \(\frac{x}{2}\sqrt{16-x^2}+8\sin^{-1}\left(\frac{x}{4}\right)+c\)
C. \(\int\frac{dx}{\sqrt{x^2-16}}\)	III. \(\frac{1}{8}\log\left|\frac{4+x}{4-x}\right|+c\)
D. \(\int\frac{dx}{16-x^2}\)	IV. \(\log\left|x+\sqrt{x^2-16}\right|+c\)

Choose the correct answer from the options given below:

• A. A-I, B-IV, C-III, D-II
• B. A-II, B-I, C-III, D-IV
• C. A-IV, B-II, C-III, D-I
• D. A-II, B-I, C-IV, D-III

Hint:

Use the standard formulas for \(\sqrt{a^2-x^2}\), \(\frac{1}{\sqrt{a^2-x^2}}\), \(\frac{1}{\sqrt{x^2-a^2}}\) and \(\frac{1}{a^2-x^2}\) with \(a=4\).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
These are the standard integrals of the form \(\sqrt{a^2-x^2}\), \(\frac{1}{\sqrt{a^2-x^2}}\), \(\frac{1}{\sqrt{x^2-a^2}}\) and \(\frac{1}{a^2-x^2}\) with \(a=4\). We match each one with its known result.

Step 2: Match A:
\[ \int\sqrt{a^2-x^2}\,dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\frac{x}{a}+c \] With \(a=4\), \(\frac{a^2}{2}=8\). This is II, so A matches II.

Step 3: Match B:
\[ \int\frac{dx}{\sqrt{a^2-x^2}}=\sin^{-1}\frac{x}{a}+c \] This is I, so B matches I.

Step 4: Match C:
\[ \int\frac{dx}{\sqrt{x^2-a^2}}=\log\left|x+\sqrt{x^2-a^2}\right|+c \] This is IV, so C matches IV.

Step 5: Match D:
\[ \int\frac{dx}{a^2-x^2}=\frac{1}{2a}\log\left|\frac{a+x}{a-x}\right|+c \] With \(a=4\), \(2a=8\). This is III, so D matches III.

Step 6: Pick the option:
We got A-II, B-I, C-IV, D-III. Option 1 has A-I which is wrong. Option 2 has C-III which is wrong. Option 3 has A-IV which is wrong. Only option 4 fits.

Final Answer:
The matching is A-II, B-I, C-IV, D-III, option 4. \[ \boxed{\text{A-II, B-I, C-IV, D-III}} \]