Use the substitution formula to evaluate the integral. $$ \int_{\pi / 3}^{2 \pi} 4 \cos ^{3} x \sin x d x $$ $$ \frac{15}{16} $$ $$ -\frac{15}{16} $$ $$ -\frac{15}{4} $$ $$ -\frac{32769}{524288} $$

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Let $$ u = \cos x $$. Then,  
$$
\frac{du}{dx} = -\sin x \quad \Longrightarrow \quad du = -\sin x\,dx \quad \Longrightarrow \quad -du = \sin x\,dx.
$$

Substitute into the integral:
$$
\int_{\pi/3}^{2\pi} 4 \cos^3 x\,\sin x\,dx = 4 \int_{\pi/3}^{2\pi} \cos^3 x\,\sin x\,dx.
$$
Replacing $$ \cos x $$ by $$ u $$ and $$ \sin x\,dx $$ by $$ -du $$, we have:
$$
4 \int_{x=\pi/3}^{x=2\pi} u^3 (\sin x\,dx) = 4 \int_{x=\pi/3}^{x=2\pi} u^3 (-du) = -4 \int_{u(\pi/3)}^{u(2\pi)} u^3\,du.
$$

Now, we need to change the limits:
$$
\text{When } x = \frac{\pi}{3}, \quad u = \cos\frac{\pi}{3} = \frac{1}{2}.
$$
$$
\text{When } x = 2\pi, \quad u = \cos 2\pi = 1.
$$

Thus, the integral becomes:
$$
-4 \int_{1/2}^{1} u^3\,du.
$$

Now, evaluate the integral:
$$
\int u^3\,du = \frac{u^4}{4}.
$$

So,
$$
-4 \int_{1/2}^{1} u^3\,du = -4 \left[\frac{u^4}{4}\right]_{1/2}^{1} = -\left[u^4\right]_{1/2}^{1} = -\left(1^4 - \left(\frac{1}{2}\right)^4\right).
$$

Simplify:
$$
1^4 = 1, \quad \left(\frac{1}{2}\right)^4 = \frac{1}{16}.
$$
$$
-\left(1 - \frac{1}{16}\right) = -\left(\frac{16}{16} - \frac{1}{16}\right) = -\frac{15}{16}.
$$

Thus, the evaluated integral is:
$$
-\frac{15}{16}.
$$
The integral evaluates to $$ -\frac{15}{16} $$.

Use the substitution formula to evaluate the integral.

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