$\int_{-2}^{2}\left(\left(x^{3} \cos \frac{x}{2}+\frac{1}{2}\right) \sqrt{4-x^{2}} d x\right.$

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Evaluate definite integrals using the chain rule

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\int_{-2}^{2}\left(\left(x^{3} \cos \frac{x}{2}+\frac{1}{2}\right) \sqrt{4-x^{2}} d x\right.

Identify Even Function: Notice that the function is even because replacing $x$ with $-x$ does not change the function. This means we can simplify the integral by evaluating from $0$ to $2$ and then doubling the result.
Set Up Integral: Set up the integral from $0$ to $2$ . $\int_{-2}^{2}\left(x^{3}\cos\left(\frac{x}{2}\right) + \frac{1}{2}\sqrt{4 - x^{2}}\right)dx = 2 \cdot \int_{0}^{2}\left(x^{3}\cos\left(\frac{x}{2}\right) + \frac{1}{2}\sqrt{4 - x^{2}}\right)dx$
Split Integral: Split the integral into two parts. $\newline$ $2 \int_{0}^{2}\left(x^{3}\cos\left(\frac{x}{2}\right) + \frac{1}{2}\sqrt{4 - x^{2}}\right)dx = 2 \left(\int_{0}^{2}\left(x^{3}\cos\left(\frac{x}{2}\right)\sqrt{4 - x^{2}}\right)dx + \int_{0}^{2}\left(\frac{1}{2}\sqrt{4 - x^{2}}\right)dx\right)$
Evaluate First Integral: Evaluate the first integral using substitution. Let $u = 4 - x^2$ , then $du = -2x \, dx$ .
Change Limits for $u$ : Change the limits of integration for $u$ . When $x = 0$ , $u = 4$ . When $x = 2$ , $u = 0$ .
Substitute and Change Limits: Substitute and change the integral limits. $\newline$ $\int_{0}^{2}(x^{3}\cos(\frac{x}{2})\sqrt{4 - x^{2}})dx = -\frac{1}{2} \int_{4}^{0}(\cos(\frac{x}{2})\sqrt{u}(-du))$
Evaluate Second Integral: Evaluate the second integral. The integral of $\sqrt{4 - x^2}$ is a standard integral that results in $(1/2)x\sqrt{4 - x^2} + 2 \arcsin(x/2) + C$ .
Calculate Integral from $0$ to $2$ : Calculate the second integral from $0$ to $2$ . $\int_{0}^{2}\left(\frac{1}{2}\sqrt{4 - x^{2}}\right)dx = \left[\frac{1}{2}x\sqrt{4 - x^2} + 2 \arcsin\left(\frac{x}{2}\right)\right]$ from $0$ to $2$
Simplify Second Integral Result: Plug in the limits for the second integral. $\left[\frac{1}{2}x\sqrt{4 - x^2} + 2 \arcsin\left(\frac{x}{2}\right)\right]$ from $0$ to $2$ = $\left[\frac{1}{2}\cdot 2\cdot\sqrt{4 - 2^2} + 2 \arcsin\left(\frac{2}{2}\right)\right] - \left[\frac{1}{2}\cdot 0\cdot\sqrt{4 - 0^2} + 2 \arcsin\left(\frac{0}{2}\right)\right]$
Add Results and Multiply: Simplify the result of the second integral. $[\frac{1}{2}\cdot2\cdot\sqrt{4 - 2^2} + 2 \arcsin(\frac{2}{2})] - [\frac{1}{2}\cdot0\cdot\sqrt{4 - 0^2} + 2 \arcsin(\frac{0}{2})] = 0 + 2 \cdot (\frac{\pi}{2}) - 0 - 0 = \pi$
Add Results and Multiply: Simplify the result of the second integral. $\newline$ $(\frac{1}{2})\cdot 2\cdot\sqrt{4 - 2^2} + 2 \arcsin(\frac{2}{2})$ - $(\frac{1}{2})\cdot 0\cdot\sqrt{4 - 0^2} + 2 \arcsin(\frac{0}{2})$ = $0$ + $2$ \cdot (\frac{\pi}{ $2$ }) - $0$ - $0$ = \pi Add the results of the two integrals together and multiply by $2$ . $\newline$ $2 \cdot (\text{First integral result} + \pi)$
Add Results and Multiply: Simplify the result of the second integral. $\newline$ $(\frac{1}{2})\cdot 2\cdot\sqrt{4 - 2^2} + 2 \arcsin(\frac{2}{2})$ - $(\frac{1}{2})\cdot 0\cdot\sqrt{4 - 0^2} + 2 \arcsin(\frac{0}{2})$ = $0$ + $2$ \cdot (\frac{\pi}{ $2$ }) - $0$ - $0$ = \pi Add the results of the two integrals together and multiply by $2$ . $\newline$ $2 \cdot (\text{First integral result} + \pi)$ Realize that the first integral involves a non-elementary function and cannot be solved using basic calculus techniques. Therefore, we cannot find an exact simplified answer for the first integral.