Match List-I with List-II (where $c$ is an arbitrary constant)

List-I	List-II
(A) $\int \tan x \, dx$	(I) $\log\vert \sec x + \tan x\vert + c$
(B) $\int \cot x \, dx$	(II) $\log\vert \sec x\vert + c$
(C) $\int \sec x \, dx$	(III) $\log\vert \sin x\vert + c$
(D) $\int \cosec x \, dx$	(IV) $\log\vert \cosec x - \cot x\vert + c$

Choose the correct answer from the options given below:

These are standard integral formulas for the four basic trigonometric functions.

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$\int \tan x \, dx$

Rewriting $\tan x$ as $\dfrac{\sin x}{\cos x}$:

$\int \dfrac{\sin x}{\cos x}\, dx$

Let $u = \cos x \Rightarrow du = -\sin x\, dx$

$= -\int \dfrac{du}{u}$

$= -\log|\cos x| + c$

$= \log\left|\dfrac{1}{\cos x}\right| + c$

$= \log|\sec x| + c$

$\therefore$ (A) $\longrightarrow$ (II)

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$\int \cot x \, dx$

Rewriting $\cot x$ as $\dfrac{\cos x}{\sin x}$:

$\int \dfrac{\cos x}{\sin x}\, dx$

Let $u = \sin x \Rightarrow du = \cos x\, dx$

$= \int \dfrac{du}{u}$

$= \log|\sin x| + c$

$\therefore$ (B) $\longrightarrow$ (III)

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$\int \sec x \, dx$

Multiplying numerator and denominator by $(\sec x + \tan x)$:

$\int \dfrac{\sec^2 x + \sec x \tan x}{\sec x + \tan x}\, dx$

Let $u = \sec x + \tan x \Rightarrow du = (\sec x \tan x + \sec^2 x)\, dx$

The numerator is exactly $du$:

$= \int \dfrac{du}{u}$

$= \log|\sec x + \tan x| + c$

$\therefore$ (C) $\longrightarrow$ (I)

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$\int \cosec x \, dx$

Multiplying numerator and denominator by $(\cosec x - \cot x)$:

$\int \dfrac{\cosec^2 x - \cosec x \cot x}{\cosec x - \cot x}\, dx$

Let $u = \cosec x - \cot x \Rightarrow du = (-\cosec x \cot x + \cosec^2 x)\, dx$

$= \int \dfrac{du}{u}$

$= \log|\cosec x - \cot x| + c$

$\therefore$ (D) $\longrightarrow$ (IV)

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(A) $\longrightarrow$ (II)

(B) $\longrightarrow$ (III)

(C) $\longrightarrow$ (I)

(D) $\longrightarrow$ (IV)