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Match List-I with List-II

[c is an arbitrary constant]

List-IList-II
(A) ∫sin⁡2x−cos⁡2xsin⁡2xcos⁡2x dx\int \frac{\sin^2 x - \cos^2 x}{\sin^2 x \cos^2 x}\,dx(I) tan⁡x−x+c\tan x - x + c
(B) ∫sec⁡2xcosec2x dx\int \frac{\sec^2 x}{\text{cosec}^2 x}\,dx(II) 2tan⁡x−3sec⁡x+c2\tan x - 3\sec x + c
(C) ∫sec⁡x(sec⁡x+tan⁡x) dx\int \sec x(\sec x + \tan x)\,dx(III) tan⁡x+cot⁡x+c\tan x + \cot x + c
(D) ∫2−3sin⁡xcos⁡2x dx\int \frac{2-3\sin x}{\cos^2 x}\,dx(IV) tan⁡x+sec⁡x+c\tan x + \sec x + c

Choose the correct answer from the options given below:

Solution

✅ Correct Option: 3

(A) splits into sec⁡2x−cosec2x\sec^2 x - \text{cosec}^2 x, giving tan⁡x+cot⁡x\tan x + \cot x. (B) reduces to ∫tan⁡2x dx=tan⁡x−x\int\tan^2 x\,dx = \tan x - x. (C) is ∫(sec⁡2x+sec⁡xtan⁡x)dx=tan⁡x+sec⁡x\int(\sec^2 x + \sec x\tan x)dx = \tan x + \sec x. (D) is ∫(2sec⁡2x−3sec⁡xtan⁡x)dx=2tan⁡x−3sec⁡x\int(2\sec^2 x - 3\sec x\tan x)dx = 2\tan x - 3\sec x.

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