Question: By first applying the half-angle formula $\sin(2x)=2\sin(x)\cos(x)$ to the integrand and then using the substitution $u=\sin(x)$, calculate the following indefinite integral: $\int \frac{7\cos(x)\sin^2(2x)}{8} \, dx$
1. **State the problem:** We want to calculate the indefinite integral $$\int \frac{7\cos(x)\sin^2(2x)}{8} \, dx.$$
2. **Apply the half-angle formula:** Recall that $$\sin(2x) = 2\sin(x)\cos(x).$$ Therefore, $$\sin^2(2x) = (2\sin(x)\cos(x))^2 = 4\sin^2(x)\cos^2(x).$$
3. **Rewrite the integrand:** Substitute $$\sin^2(2x)$$ with $$4\sin^2(x)\cos^2(x)$$ in the integral:
$$\int \frac{7\cos(x) \cdot 4\sin^2(x)\cos^2(x)}{8} \, dx = \int \frac{28\cos^3(x)\sin^2(x)}{8} \, dx = \int \frac{7}{2} \cos^3(x) \sin^2(x) \, dx.$$
4. **Use substitution:** Let $$u = \sin(x)$$ so that $$du = \cos(x) \, dx.$$
5. **Rewrite the integral in terms of $u$:** We have $$\cos^3(x) = \cos^2(x) \cos(x) = (1 - \sin^2(x)) \cos(x) = (1 - u^2) \cos(x).$$
Thus, the integral becomes
$$\int \frac{7}{2} (1 - u^2) u^2 \cos(x) \, dx = \frac{7}{2} \int (1 - u^2) u^2 \, du = \frac{7}{2} \int (u^2 - u^4) \, du.$$
6. **Integrate:**
$$\frac{7}{2} \left( \frac{u^3}{3} - \frac{u^5}{5} \right) + C = \frac{7}{2} \left( \frac{\sin^3(x)}{3} - \frac{\sin^5(x)}{5} \right) + C.$$
7. **Final answer:**
$$\boxed{\frac{7}{6} \sin^3(x) - \frac{7}{10} \sin^5(x) + C}.$$