Evaluate the integral. $\int x \sin 3 x \cos 3 x d x$
A) $-\frac{1}{9}(\cos 9 x-9 x \sin 9 x)+C$
B) $\frac{1}{18}(\sin 3 x-3 x \cos 3 x)+C$
C) $\frac{1}{9}(\sin 9 x-49 \cos 9 x)+C$
D) $-\frac{1}{18}(\sin 18 x-18 x \cos 18 x)+C$
E) None of these

Question

Quizplus · Accepted Answer

We use integration by parts with $u=x$ and $dv=\sin 3x\cos 3x \, dx$. Then $du=dx$ and $v= \frac{1}{2}\sin^2 3x$. Therefore, $$\int x\sin 3x \cos 3x \,dx = -\frac{x}{6}\cos 3x+\frac{1}{36}\sin^23x +C.$$ Then we use the identity $\sin^2 	heta = \frac{1-\cos 2	heta}{2}$ to get the answer $$\frac{1}{18}(\sin 3x-3x\cos 3x)+C.$$

Evaluate the Integral ∫xsin⁡3xcos⁡3xdx\int x \sin 3 x \cos 3 x d x∫xsin3xcos3xdx

Evaluate the Integral $\int x \sin 3 x \cos 3 x d x$