Evaluate the integral by using multiple substitutions.
-$$\int \frac { x \cos 6 x ^ { 2 } \sqrt { \sin 6 x ^ { 2 } } } { 5 } d x$$
A) $\frac { \sin ^ { 3 / 2 } x ^ { 2 } } { 15 } + C$
B) $\frac { \sqrt { \sin 6 x ^ { 2 } } } { 60 } + C$
C) $\frac { \sin ^ { 3 / 2 } 6 x ^ { 2 } } { 90 } + C$
D) $\frac { \sin ^ { 3 / 2 } x } { 90 } + C$

Question

Quizplus · Accepted Answer

We can use the substitution $u = 6x^2$ to simplify the integrand. Then, we use the substitution $v = \sqrt{u}$ to further simplify it. 
\begin{align*}
\int \frac { x \cos 6 x ^ { 2 } \sqrt { \sin 6 x ^ { 2 } } } { 5 } d x &= \frac{1}{5} \int \cos u \cdot \sin v\, dv \
&= \frac{1}{5} \left(-\cos u \cdot \cos v + \int \sin u \cdot \cos v\, duight) \
&= \frac{1}{5} \left(-\cos u \cdot \cos v + \frac{1}{2}\sin u \cdot \sin vight) + C \
&= \frac{1}{5} \left(-\cos 6x^2 \cdot \sqrt{6x^2} + \frac{1}{2}\sin 6x^2 \cdot \sqrt{6x^2}ight) + C \
&= \frac{1}{5} \left(-\sqrt{6}x\cos 6x^2 + \frac{\sqrt{6}}{2}\sin 6x^2ight) + C\
&= \frac{\sin ^ { 3 / 2 } 6 x ^ { 2 } } { 90 } + C
\end{align*}
Therefore, the answer is (C).

Evaluate the Integral by Using Multiple Substitutions ∫xcos⁡6x2sin⁡6x25dx\int \frac { x \cos 6 x ^ { 2 } \sqrt { \sin 6 x ^ { 2 } } } { 5 } d x∫5xcos6x2sin6x2​​dx

Evaluate the Integral by Using Multiple Substitutions $\int \frac { x \cos 6 x ^ { 2 } \sqrt { \sin 6 x ^ { 2 } } } { 5 } d x$