Integrate $\int \cos ^{2} 3 x \sin 3 x d x$ .
A) $\frac{\cos ^{3} 3 x}{9}+C$
B) $-\frac{\cos ^{3} 3 x}{9}+C$
C) $\frac{\cos ^{3} 3 x}{3}+C$
D) $-\frac{\cos ^{3} 3 x}{27}+C$

Question

Quizplus · Accepted Answer

The integral $\int \cos^2 3x \sin 3x dx$ can be solved using the substitution method. Let $u = \cos 3x$, then $du = -3 \sin 3x dx$, or $-\frac{1}{3} du = \sin 3x dx$. Substituting these into the integral gives $-\frac{1}{3} \int u^2 du$, which integrates to $-\frac{1}{3} \cdot \frac{u^3}{3} + C = -\frac{\cos^3 3x}{9} + C$.

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