Consider the following statements:
(A) \(\sin^2x\) is an anti-derivative of \(\sin 2x\).
(B) \(-\cos^2x\) is an anti-derivative of \(\sin 2x\).
(C) \(\displaystyle -\frac{1}{2}\cos 2x\) is an anti-derivative of \(\sin 2x\).
(D) \(\cos^2x\) is an anti-derivative of \(\sin 2x\).
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A), (B) and (C) only
A function \(F\) is an anti-derivative of \(\sin 2x\) when \(F'(x) = \sin 2x\). Two anti-derivatives may look different and still differ only by a constant.
\( \displaystyle \frac{d}{dx}\sin^2x = 2\sin x\cos x = \sin 2x \)
\( \displaystyle \frac{d}{dx}\left(-\cos^2x\right) = 2\cos x\sin x = \sin 2x \)
\( \displaystyle \frac{d}{dx}\left(-\frac{1}{2}\cos 2x\right) = \sin 2x \)
\( \displaystyle \frac{d}{dx}\cos^2x = -2\cos x\sin x = -\sin 2x \)
So \(\sin^2x\), \(-\cos^2x\) and \(-\frac{1}{2}\cos 2x\) all have derivative \(\sin 2x\) (they differ by constants, since \(\sin^2x = -\cos^2x+1\)), while \(\cos^2x\) has derivative \(-\sin 2x\).
Checking the statements
(A) \(\sin^2x\) is an anti-derivative of \(\sin 2x\). — Correct. Its derivative is \(2\sin x\cos x = \sin 2x\).
(B) \(-\cos^2x\) is an anti-derivative of \(\sin 2x\). — Correct. Its derivative is \(2\cos x\sin x = \sin 2x\).
(C) \(-\frac{1}{2}\cos 2x\) is an anti-derivative of \(\sin 2x\). — Correct. Its derivative is \(-\frac{1}{2}\cdot(-2\sin 2x) = \sin 2x\).
(D) \(\cos^2x\) is an anti-derivative of \(\sin 2x\). — Incorrect. Its derivative is \(-2\cos x\sin x = -\sin 2x\).