Evaluate the integral.
-$$\int 4 \cos ^ { 4 } 6 x d x$$
A) $3 x + \frac { 1 } { 6 } \sin 6 x + \frac { 1 } { 12 } \sin 24 x + C$
B) $3 x + \frac { 1 } { 6 } \sin 6 x + \frac { 1 } { 48 } \sin 12 x + C$
C) $\frac { 3 } { 2 } x + \frac { 1 } { 3 } \sin 12 x + \frac { 1 } { 8 } \sin 24 x + C$
D) $\frac { 3 } { 2 } x + \frac { 1 } { 6 } \sin 12 x + \frac { 1 } { 48 } \sin 24 x + C$

Question

Quizplus · Accepted Answer

We can use the identity $\cos^4x=\frac{1}{8}(3+\cos 4x+4\cos 2x)$ to rewrite the integral as 
\begin{align*}
\int 4 \cos ^ { 4 } 6 x d x &= \int \frac{1}{2}(3+\cos 24x+4\cos 12x)\,dx \
&= \frac{3}{2}x+\frac{1}{6}\sin 12x+\frac{1}{48}\sin 24x+C
\end{align*}
Thus, the answer is $\boxed{	extbf{(D)}}$.

Evaluate the Integral ∫4cos⁡46xdx\int 4 \cos ^ { 4 } 6 x d x∫4cos46xdx

Evaluate the Integral $\int 4 \cos ^ { 4 } 6 x d x$