Question

If \(\int \sin ^{3}x\cdot \cos ^{3}xdx=\frac{1}{32}\left[\frac{3}{2}f(x)+\frac{1}{6}g(x)\right]+C\), then \(f(x)\) and \(g(x)\) are respectively...

• A. \(-\sin 2x,\,\sin 6x\)
• B. \(-\cos 2x,\,\cos 6x\)
• C. \(\cos 6x,-\cos 2x\)
• D. \(\sin 2x,\,\sin 6x\)

Hint:

Write sin x cos x as half of sin 2x, expand the cube and use the triple angle identity before integrating.

Answer 1

The Correct Option is B

Step 1: Rewrite the integrand
$\sin^3x\cos^3x=\left(\dfrac{\sin2x}2\right)^3=\dfrac{\sin^32x}8$.

Step 2: Reduce the power
$\sin^32x=\dfrac34\sin2x-\dfrac14\sin6x$, so the integrand is $\dfrac3{32}\sin2x-\dfrac1{32}\sin6x$.

Step 3: Integrate term by term
$\dfrac{3}{32}\cdot\dfrac{-\cos2x}{2}-\dfrac1{32}\cdot\dfrac{-\cos6x}{6}=\dfrac1{32}\left[\dfrac32(-\cos2x)+\dfrac16\cos6x\right]$.

Final Answer:
$f(x)=-\cos2x$ and $g(x)=\cos6x$. \[ \boxed{\text{(B) }-\cos 2x,\,\cos 6x} \]