Use reduction formulas to evaluate the integral.
-$\int \sin ^ { 2 } 3 x \cos ^ { 2 } 3 x d x$
A) $- \frac { 1 } { 3 } \sin 3 x \cos ^ { 3 } 3 x + \frac { x } { 8 } + \frac { 1 } { 3 } \sin 6 x + C$
B) $- \frac { 1 } { 12 } \sin 3 x \cos ^ { 2 } 3 x + \frac { x } { 8 } + \frac { 1 } { 48 } \sin 6 x + C$
C) $- \frac { 1 } { 12 } \sin 3 x \cos ^ { 3 } 3 x + \frac { x } { 8 } + \frac { 1 } { 48 } \sin 6 x + C$
D) $- \frac { 1 } { 12 } \sin 3 x \cos ^ { 3 } 3 x + \frac { x } { 8 } + \frac { 1 } { 48 } \cos 6 x + C$

Question

Quizplus · Accepted Answer

We can use the reduction formula for $\sin^2$ to rewrite $\sin^2 3x$ as $\frac{1}{2}(1-\cos 6x)$ and use the reduction formula for $\cos^2$ to rewrite $\cos^2 3x$ as $\frac{1}{2}(1+\cos 6x)$. Thus,
\begin{align*}
\int \sin ^ { 2 } 3 x \cos ^ { 2 } 3 x d x &= \int \frac{1}{4}(1-\cos 6x)(1+\cos 6x) dx \
&= \int \frac{1}{4}(1-\cos^2 6x) dx \
&= \int \frac{1}{8}(2-\cos^2 6x) dx \
&= \frac{1}{8}(2x-\frac{1}{2}(x+\frac{1}{6}\sin 6x)) + C \
&= \frac{1}{12}\sin 6x + \frac{1}{8}x - \frac{1}{16}\cos^2 3x + C.
\end{align*}
Using the reduction formula for $\cos^2$, we have $\int \cos^2 3x dx = \frac{1}{2}x+\frac{1}{4}\sin 6x + C$, and using this formula gives us the answer $\boxed{	extbf{(C)}}$.

Use Reduction Formulas to Evaluate the Integral ∫sin⁡23xcos⁡23xdx\int \sin ^ { 2 } 3 x \cos ^ { 2 } 3 x d x∫sin23xcos23xdx

Use Reduction Formulas to Evaluate the Integral $\int \sin ^ { 2 } 3 x \cos ^ { 2 } 3 x d x$