Consider the following statements, where C is a constant of integration:
(A) \(\int\sin^3x\cos x\,dx = \displaystyle\frac{\sin^4x}{4}+C\)
(B) \(\int e^{\tan x}\sec^2x\,dx = e^{\tan x}+C\)
(C) \(\int\displaystyle\frac{\sin x}{1+\cos x}\,dx = \log|1+\cos x|+C\)
(D) \(\int\displaystyle\frac{2x}{1+x^4}\,dx = \frac{1}{2}\tan^{-1}\left(x^2\right)+C\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) and (B) only
Each integral is found by a substitution t = the inner function:
\( \displaystyle t = \sin x,\ dt = \cos x\,dx:\quad \int t^3\,dt = \frac{t^4}{4} \)
\(t = \tan x,\ dt = \sec^2x\,dx:\quad \int e^t\,dt = e^t\)
\( \displaystyle t = 1+\cos x,\ dt = -\sin x\,dx:\quad -\int\frac{dt}{t} = -\log|t| \)
\( \displaystyle t = x^2,\ dt = 2x\,dx:\quad \int\frac{dt}{1+t^2} = \tan^{-1}t \)
So the integrals are \(\frac{\sin^4x}{4}\), \(e^{\tan x}\), \(-\log|1+\cos x|\) and \(\tan^{-1}\left(x^2\right)\), each plus C.
Checking the statements
(A) \(\int\sin^3x\cos x\,dx = \frac{\sin^4x}{4}+C\) — Correct. With \(t = \sin x\) it is \(\int t^3\,dt = \frac{t^4}{4}\).
(B) \(\int e^{\tan x}\sec^2x\,dx = e^{\tan x}+C\) — Correct. With \(t = \tan x\) it is \(\int e^t\,dt = e^t\).
(C) \(\int\frac{\sin x}{1+\cos x}\,dx = \log|1+\cos x|+C\) — Incorrect. The derivative of \(1+\cos x\) is \(-\sin x\), so the answer is \(-\log|1+\cos x|+C\).
(D) \(\int\frac{2x}{1+x^4}\,dx = \frac{1}{2}\tan^{-1}\left(x^2\right)+C\) — Incorrect. The numerator 2x is exactly dt for \(t = x^2\), so no factor \(\frac{1}{2}\) appears; the answer is \(\tan^{-1}\left(x^2\right)+C\).