Evaluate the integral by using multiple substitutions.
-$\int \frac { 3 \sin ^ { 2 } \sqrt { 3 y + 9 } \cos \sqrt { 3 y + 9 } } { 2 \sqrt { 3 y + 9 } } d y$
A) $2 \sin \sqrt { 3 y + 9 } + C$
B) $\frac { \sin ^ { 3 } \sqrt { 3 y + 9 } } { 18 } + C$
C) $\frac { \sin ^ { 3 } \sqrt { 3 y + 9 } } { 3 } + C$
D) $\frac { \sqrt { 3 y + 9 ^ { 3 } } } { 3 } + C$

Question

Quizplus · Accepted Answer

Let's make the substitution $u=\sqrt{3y+9}.$ Then we have $du=\frac{3}{2\sqrt{3y+9}}dy.$ 
Making the substitution in our integral, we get $$\int \frac { 3 \sin ^ { 2 } \sqrt { 3 y + 9 } \cos \sqrt { 3 y + 9 } } { 2 \sqrt { 3 y + 9 } } d y = \frac{2}{3} \int \sin^2u\cos u du.$$ 
Let's make the substitution $v=\sin u,$ so that $dv=\cos u du.$ We can rewrite our integral as 
\begin{align*}
\frac{2}{3}\int \sin^2u\cos u du &= \frac{2}{3}\int v^2dv \
&= \frac{2}{9}v^3 + C \
&= \frac{2}{9}(\sin u)^3 + C. 
\end{align*} 
Making the substitution $u=\sqrt{3y+9},$ we have the final answer: $\boxed{	extbf{(C)}\ \frac {\sin ^ { 3 } \sqrt { 3 y + 9 } } { 3 } + C}.$

Evaluate the Integral by Using Multiple Substitutions ∫3sin⁡23y+9cos⁡3y+923y+9dy\int \frac { 3 \sin ^ { 2 } \sqrt { 3 y + 9 } \cos \sqrt { 3 y + 9 } } { 2 \sqrt { 3 y + 9 } } d y∫23y+9​3sin23y+9​cos3y+9​​dy

Evaluate the Integral by Using Multiple Substitutions $\int \frac { 3 \sin ^ { 2 } \sqrt { 3 y + 9 } \cos \sqrt { 3 y + 9 } } { 2 \sqrt { 3 y + 9 } } d y$