Question

\(\int \sin ^{3}x\cos ^{3}xdx=\ldots\)

• A. \(-\frac{1}{192}[9\,\cos 2x-\cos 6x]+c\)
• B. \(\frac{1}{192}[9\,\cos 2x-\cos 6x]+c\)
• C. \(-\frac{1}{32}[\cos 2x-\cos 6x]+c\)
• D. \(\frac{1}{32}[\cos 2x-\cos 6x]+c\)

Hint:

Write sin x cos x as half of sin 2x, then use the triple-angle formula to reduce the cube of sine.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
Use \(\sin x\cos x=\frac{\sin2x}{2}\) so that the integrand becomes \(\frac{\sin^{3}2x}{8}\).

Step 2: Reduce the power
Using \(\sin^{3}u=\frac{3\sin u-\sin3u}{4}\) with \(u=2x\):
\(\sin^{3}x\cos^{3}x=\frac{3\sin2x-\sin6x}{32}\).

Step 3: Integrate
\(\int\sin2x\,dx=-\frac{\cos2x}{2}\) and \(\int\sin6x\,dx=-\frac{\cos6x}{6}\).
So the integral is \(\frac{1}{32}\left[-\frac{3\cos2x}{2}+\frac{\cos6x}{6}\right]+c\).

Step 4: Simplify
\(=\frac{1}{192}\left[-9\cos2x+\cos6x\right]+c=-\frac{1}{192}[9\cos2x-\cos6x]+c\).

Final Answer:
The integral is −(1/192)[9 cos 2x − cos 6x] + c. \[ \boxed{\text{(A) }-\dfrac{1}{192}[9\cos 2x-\cos 6x]+c} \]