Consider the following statements, where C is a constant of integration:
(A) \(\int\cos^2x\,dx = \displaystyle\frac{x}{2}-\frac{\sin 2x}{4}+C\)
(B) \(\int\sin^3x\,dx = \displaystyle -\cos x+\frac{\cos^3x}{3}+C\)
(C) \(\int\sin^2x\cos^2x\,dx = \displaystyle\frac{x}{8}-\frac{\sin 4x}{32}+C\)
(D) \(\int\sqrt{1+\sin 2x}\,dx = \sin x-\cos x+C\), for \(\displaystyle 0 \lt x \lt \frac{\pi}{2}\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (B), (C) and (D) only
Each integrand is first rewritten with a trigonometric identity:
\( \displaystyle \cos^2x = \frac{1+\cos 2x}{2},\quad \int\cos^2x\,dx = \frac{x}{2}+\frac{\sin 2x}{4} \)
\( \sin^3x = \left(1-\cos^2x\right)\sin x \)
\( \displaystyle t = \cos x:\quad -\int\left(1-t^2\right)dt = -\cos x+\frac{\cos^3x}{3} \)
\( \displaystyle \sin^2x\cos^2x = \frac{\sin^2 2x}{4} = \frac{1-\cos 4x}{8} \)
\( \displaystyle \int\frac{1-\cos 4x}{8}\,dx = \frac{x}{8}-\frac{\sin 4x}{32} \)
\( 1+\sin 2x = \left(\sin x+\cos x\right)^2 \)
\( \displaystyle \int\left(\sin x+\cos x\right)dx = \sin x-\cos x \)
So \(\int\cos^2x\,dx = \frac{x}{2}+\frac{\sin 2x}{4}+C\), while (B), (C) and (D) are true (in (D), \(\sin x+\cos x \gt 0\) on the given interval).
Checking the statements
(A) \(\int\cos^2x\,dx = \frac{x}{2}-\frac{\sin 2x}{4}+C\) — Incorrect. The sine term is \(+\frac{\sin 2x}{4}\); the given result is the integral of \(\sin^2x\).
(B) \(\int\sin^3x\,dx = -\cos x+\frac{\cos^3x}{3}+C\) — Correct. Its derivative is \(\sin x-\cos^2x\sin x = \sin^3x\).
(C) \(\int\sin^2x\cos^2x\,dx = \frac{x}{8}-\frac{\sin 4x}{32}+C\) — Correct. Its derivative is \(\frac{1-\cos 4x}{8} = \sin^2x\cos^2x\).
(D) \(\int\sqrt{1+\sin 2x}\,dx = \sin x-\cos x+C\) — Correct. Its derivative is \(\cos x+\sin x = \sqrt{1+\sin 2x}\) there.