Question

Evaluate: $\int \sin ^{3} x \cos ^{3} x d x$

• A. $\frac{1}{32}\left\{\frac{3}{2} \cos 2 x-\frac{1}{6} \cos 6 x\right\}+C$
• B. $\frac{1}{32}\left\{-\frac{3}{2} \cos 2 x+\frac{1}{6} \cos 6 x\right\}+C$
• C. $\frac{1}{32}\left\{-\frac{3}{2} \cos 2 x-\frac{1}{6} \cos 6 x\right\}+C$
• D. None of these

Answer 1

Answer: (B)

Let $I =\int \sin ^{3} x \cos ^{3} x\, dx$. Then,
$I = \frac{1}{8} \int(2 \sin x \cos x)^{3} d x$
$\Rightarrow I =\frac{1}{8} \int \sin ^{3} 2 x\, dx$
$\Rightarrow I =\frac{1}{8} \int \frac{3 \sin 2 x -\sin 6 x }{4} d x$
$\Rightarrow I =\frac{1}{32} \int(3 \sin 2 x -\sin 6 x ) d x$
$=\frac{1}{32}\left\{-\frac{3}{2} \cos 2 x +\frac{1}{6} \cos 6 x \right\}+ C$