Find the indefinite integral. (Use $$ C $$ for the constant of integration.) $$ \int \sin ^{3}(4 \theta) \sqrt{\cos (4 \theta)} d \theta $$

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We begin with the integral
$$
\int \sin^{3}(4\theta)\sqrt{\cos(4\theta)}\,d\theta.
$$
Notice that the integrand contains both $$\sin(4\theta)$$ and $$\cos(4\theta)$$. To simplify the expression, write $$\sin^3(4\theta)$$ as
$$
\sin^3(4\theta) = \sin(4\theta)\sin^2(4\theta)
= \sin(4\theta)\left(1-\cos^2(4\theta)\right).
$$
Thus the integral becomes
$$
\int \sin(4\theta)(1-\cos^2(4\theta))\sqrt{\cos(4\theta)}\,d\theta.
$$

Let
$$
u=\cos(4\theta).
$$
Then the derivative is
$$
\frac{du}{d\theta} = -4\sin(4\theta) \quad \Longrightarrow \quad \sin(4\theta)\,d\theta = -\frac{1}{4}\,du.
$$
Substitute into the integral:
$$
\int \sin(4\theta)(1-u^2)\sqrt{u}\,d\theta = -\frac{1}{4}\int (1-u^2)\sqrt{u}\,du.
$$

Expand the integrand:
$$
(1-u^2)\sqrt{u} = u^{1/2} - u^{5/2}.
$$
So the integral becomes
$$
-\frac{1}{4}\int \left(u^{1/2}-u^{5/2}\right)du.
$$

Integrate term‐by‐term:
$$
\int u^{1/2}\,du = \frac{2}{3}u^{3/2} \quad \text{and} \quad \int u^{5/2}\,du = \frac{2}{7}u^{7/2}.
$$
Thus,
$$
-\frac{1}{4}\left[\frac{2}{3}u^{3/2}-\frac{2}{7}u^{7/2}\right] + C.
$$
Simplify by factoring the constant:
$$
-\frac{1}{4}\left[\frac{2}{3}u^{3/2}-\frac{2}{7}u^{7/2}\right] = -\frac{1}{6}u^{3/2} + \frac{1}{14}u^{7/2}.
$$

Returning to the variable $$\theta$$, recall that $$u=\cos(4\theta)$$. The final answer is
$$
-\frac{1}{6}\cos^{3/2}(4\theta)+\frac{1}{14}\cos^{7/2}(4\theta)+C.
$$
The indefinite integral is: $$ -\frac{1}{6}\cos^{3/2}(4\theta) + \frac{1}{14}\cos^{7/2}(4\theta) + C $$

Find the indefinite integral. (Use for the constant of integration.)

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