Find: integral of sin³(x)cos(x/2)

To find the integral of $\sin^3(x) \cos\left(\frac{x}{2}\right)$, we can use a substitution. Let: $$u = \sin(x)$$ $$du = \cos(x) dx$$ Now, we can rewrite the integral as: $$\int u^3 \cos\left(\frac{x}{2}\right) du$$ Since $du = \cos(x) dx$, we need to express $\cos(x)$ in terms of $u$. From the trigonometric identity $\sin^2(x) + \cos^2(x) = 1$, we have: $$\cos^2(x) = 1 - \sin^2(x)$$ $$\cos(x) = \sqrt{1 - u^2}$$ Now, substitute $\cos(x) = \sqrt{1 - u^2}$ into the integral: $$\int u^3 \sqrt{1 - u^2} du$$ This integral can be solved using trigonometric substitution. Let: $$u = \sin(\theta)$$ $$du = \cos(\theta) d\theta$$ Now, rewrite the integral in terms of $\theta$: $$\int \sin^3(\theta) \sqrt{1 - \sin^2(\theta)} \cos(\theta) d\theta$$ $$\int \sin^3(\theta) \sqrt{\cos^2(\theta)} \cos(\theta) d\theta$$ $$\int \sin^3(\theta) \cos^2(\theta) d\theta$$ $$\int \sin^3(\theta) (1 - \sin^2(\theta)) d\theta$$ $$\int (\sin^3(\theta) - \sin^5(\theta)) d\theta$$ This integral can be solved using standard trigonometric integral formulas. After integrating, don't forget to revert back to the original variable $x$ using the original substitution.