Match List-I with List-II (C is a constant of integration)
| List-I | List-II | ||
| (A) | \(\int\displaystyle\frac{dx}{\sin x\cos x}\) | (I) | \(\tan x-x+C\) |
| (B) | \(\int\displaystyle\frac{\cos 2x}{\sin^2x\cos^2x}\,dx\) | (II) | \(\displaystyle -\frac{1}{4}\cos 4x-\frac{1}{2}\cos 2x+C\) |
| (C) | \(\int\displaystyle\frac{1-\cos 2x}{1+\cos 2x}\,dx\) | (III) | \(-\cot x-\tan x+C\) |
| (D) | \(\int 2\sin 3x\cos x\,dx\) | (IV) | \(\log|\tan x|+C\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(IV), (B)-(III), (C)-(I), (D)-(II)
(A) \(\int\frac{dx}{\sin x\cos x}\)
This matches (IV) \(\log|\tan x|+C\).
Dividing numerator and denominator by \(\cos^2x\):
\( \displaystyle \frac{1}{\sin x\cos x} = \frac{\sec^2x}{\tan x} \)
With \(t = \tan x\), \(dt = \sec^2x\,dx\), the integral is \(\int\frac{dt}{t} = \log|\tan x|+C\).
(B) \(\int\frac{\cos 2x}{\sin^2x\cos^2x}\,dx\)
This matches (III) \(-\cot x-\tan x+C\).
Using \(\cos 2x = \cos^2x-\sin^2x\):
\( \displaystyle \frac{\cos^2x-\sin^2x}{\sin^2x\cos^2x} = \mathrm{cosec}^2x-\sec^2x \)
\( \displaystyle \int\left(\mathrm{cosec}^2x-\sec^2x\right)dx = -\cot x-\tan x+C \)
(C) \(\int\frac{1-\cos 2x}{1+\cos 2x}\,dx\)
This matches (I) \(\tan x-x+C\).
Using \(1-\cos 2x = 2\sin^2x\) and \(1+\cos 2x = 2\cos^2x\):
\( \displaystyle \frac{1-\cos 2x}{1+\cos 2x} = \tan^2x = \sec^2x-1 \)
\( \displaystyle \int\left(\sec^2x-1\right)dx = \tan x-x+C \)
(D) \(\int 2\sin 3x\cos x\,dx\)
This matches (II) \(-\frac{1}{4}\cos 4x-\frac{1}{2}\cos 2x+C\).
Using the product-to-sum formula \(2\sin A\cos B = \sin(A+B)+\sin(A-B)\):
\(2\sin 3x\cos x = \sin 4x+\sin 2x\)
\( \displaystyle \int(\sin 4x+\sin 2x)\,dx = -\frac{1}{4}\cos 4x-\frac{1}{2}\cos 2x+C \)