Match List-I with List-II (C is a constant of integration)
| List-I | List-II | ||
| (A) | \(\int\sec^2 3x\,dx\) | (I) | \(\displaystyle\frac{1}{3}\sin^{-1}(3x)+C\) |
| (B) | \(\int\displaystyle\frac{dx}{1+9x^2}\) | (II) | \(\displaystyle\frac{1}{3}\log|\sec 3x|+C\) |
| (C) | \(\int\displaystyle\frac{dx}{\sqrt{1-9x^2}}\) | (III) | \(\displaystyle\frac{1}{3}\tan 3x+C\) |
| (D) | \(\int\tan 3x\,dx\) | (IV) | \(\displaystyle\frac{1}{3}\tan^{-1}(3x)+C\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
Each integral is a standard integral with x replaced by 3x, so each result carries the factor \(\frac{1}{3}\).
(A) \(\int\sec^2 3x\,dx\)
This matches (III) \(\frac{1}{3}\tan 3x+C\).
\( \displaystyle \frac{d}{dx}\left(\frac{1}{3}\tan 3x\right) = \frac{1}{3}\times 3\sec^2 3x = \sec^2 3x \)
(B) \(\int\frac{dx}{1+9x^2}\)
This matches (IV) \(\frac{1}{3}\tan^{-1}(3x)+C\).
With \(t = 3x\), \(dx = \frac{dt}{3}\):
\( \displaystyle \frac{1}{3}\int\frac{dt}{1+t^2} = \frac{1}{3}\tan^{-1}(3x)+C \)
(C) \(\int\frac{dx}{\sqrt{1-9x^2}}\)
This matches (I) \(\frac{1}{3}\sin^{-1}(3x)+C\).
With \(t = 3x\):
\( \displaystyle \frac{1}{3}\int\frac{dt}{\sqrt{1-t^2}} = \frac{1}{3}\sin^{-1}(3x)+C \)
(D) \(\int\tan 3x\,dx\)
This matches (II) \(\frac{1}{3}\log|\sec 3x|+C\).
Using \(\int\tan t\,dt = \log|\sec t|\) with \(t = 3x\):
\( \displaystyle \int\tan 3x\,dx = \frac{1}{3}\log|\sec 3x|+C \)