Consider the following statements, where C is a constant of integration:
(A) \(\int\sec x\tan x\,dx = \sec x+C\)
(B) \(\int\mathrm{cosec}^2x\,dx = \cot x+C\)
(C) \(\int\displaystyle\frac{dx}{1+x^2} = \tan^{-1}x+C\)
(D) \(\int\displaystyle\frac{dx}{\sqrt{1-x^2}} = \cos^{-1}x+C\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A) and (C) only
Integration is the reverse of differentiation, so each result is checked by differentiating the right-hand side:
\( \displaystyle \frac{d}{dx}\sec x = \sec x\tan x \)
\( \displaystyle \frac{d}{dx}\cot x = -\mathrm{cosec}^2x \)
\( \displaystyle \frac{d}{dx}\tan^{-1}x = \frac{1}{1+x^2} \)
\( \displaystyle \frac{d}{dx}\cos^{-1}x = -\frac{1}{\sqrt{1-x^2}} \)
So \(\int\mathrm{cosec}^2x\,dx = -\cot x+C\) and \(\int\frac{dx}{\sqrt{1-x^2}} = \sin^{-1}x+C\), while (A) and (C) are the standard results.
Checking the statements
(A) \(\int\sec x\tan x\,dx = \sec x+C\) — Correct. The derivative of \(\sec x\) is \(\sec x\tan x\).
(B) \(\int\mathrm{cosec}^2x\,dx = \cot x+C\) — Incorrect. The derivative of \(\cot x\) is \(-\mathrm{cosec}^2x\), so the answer is \(-\cot x+C\).
(C) \(\int\frac{dx}{1+x^2} = \tan^{-1}x+C\) — Correct. The derivative of \(\tan^{-1}x\) is \(\frac{1}{1+x^2}\).
(D) \(\int\frac{dx}{\sqrt{1-x^2}} = \cos^{-1}x+C\) — Incorrect. The derivative of \(\cos^{-1}x\) carries a minus sign; the answer is \(\sin^{-1}x+C\).