Key Concepts and Formulas

• Integration by parts: $\int u \, dv = uv - \int v \, du$
• Derivatives of trigonometric functions: $\frac{d}{dx}(\sin x) = \cos x$, $\frac{d}{dx}(\cos x) = -\sin x$
• Derivatives of polynomial functions: $\frac{d}{dx}(x^n) = nx^{n-1}$

Step-by-Step Solution

Step 1: Integrate $\int x^3 \sin x \, dx$ using integration by parts.

Let $u = x^3$ and $dv = \sin x \, dx$. Then $du = 3x^2 \, dx$ and $v = -\cos x$. Applying integration by parts: $\int x^3 \sin x \, dx = -x^3 \cos x - \int (-\cos x)(3x^2) \, dx = -x^3 \cos x + 3\int x^2 \cos x \, dx$ We have reduced the power of $x$ inside the integral.

Step 2: Integrate $\int x^2 \cos x \, dx$ using integration by parts.

Let $u = x^2$ and $dv = \cos x \, dx$. Then $du = 2x \, dx$ and $v = \sin x$. Applying integration by parts: $\int x^2 \cos x \, dx = x^2 \sin x - \int (\sin x)(2x) \, dx = x^2 \sin x - 2\int x \sin x \, dx$

Step 3: Integrate $\int x \sin x \, dx$ using integration by parts.

Let $u = x$ and $dv = \sin x \, dx$. Then $du = dx$ and $v = -\cos x$. Applying integration by parts: $\int x \sin x \, dx = -x \cos x - \int (-\cos x) \, dx = -x \cos x + \int \cos x \, dx = -x \cos x + \sin x$

Step 4: Substitute the results back into the original integral.

Substituting the result from Step 3 into Step 2: $\int x^2 \cos x \, dx = x^2 \sin x - 2(-x \cos x + \sin x) = x^2 \sin x + 2x \cos x - 2 \sin x$ Substituting this into the result from Step 1: $\int x^3 \sin x \, dx = -x^3 \cos x + 3(x^2 \sin x + 2x \cos x - 2 \sin x) + C = -x^3 \cos x + 3x^2 \sin x + 6x \cos x - 6 \sin x + C$

Step 5: Identify $g(x)$ and find $g(\frac{\pi}{2})$.

From the problem statement, we have $g(x) = -x^3 \cos x + 3x^2 \sin x + 6x \cos x - 6 \sin x$. Then, $g\left(\frac{\pi}{2}\right) = -\left(\frac{\pi}{2}\right)^3 \cos\left(\frac{\pi}{2}\right) + 3\left(\frac{\pi}{2}\right)^2 \sin\left(\frac{\pi}{2}\right) + 6\left(\frac{\pi}{2}\right) \cos\left(\frac{\pi}{2}\right) - 6 \sin\left(\frac{\pi}{2}\right) = 0 + 3\left(\frac{\pi^2}{4}\right)(1) + 0 - 6(1) = \frac{3\pi^2}{4} - 6$

Step 6: Find $g'(x)$ and $g'(\frac{\pi}{2})$.

$g'(x) = \frac{d}{dx} \left( -x^3 \cos x + 3x^2 \sin x + 6x \cos x - 6 \sin x \right)$ $g'(x) = -3x^2 \cos x + x^3 \sin x + 6x \sin x + 3x^2 \cos x + 6 \cos x - 6x \sin x - 6 \cos x$ $g'(x) = x^3 \sin x$ Then, $g'\left(\frac{\pi}{2}\right) = \left(\frac{\pi}{2}\right)^3 \sin\left(\frac{\pi}{2}\right) = \frac{\pi^3}{8} (1) = \frac{\pi^3}{8}$

Step 7: Calculate $8\left(g\left(\frac{\pi}{2}\right)+g^{\prime}\left(\frac{\pi}{2}\right)\right)$.

$8\left(g\left(\frac{\pi}{2}\right)+g^{\prime}\left(\frac{\pi}{2}\right)\right) = 8\left(\frac{3\pi^2}{4} - 6 + \frac{\pi^3}{8}\right) = 8\left(\frac{6\pi^2 - 48 + \pi^3}{8}\right) = \pi^3 + 6\pi^2 - 48$

Step 8: Determine $\alpha$, $\beta$, and $\gamma$.

We have $\pi^3 + 6\pi^2 - 48 = \alpha \pi^3 + \beta \pi^2 + \gamma$. Therefore, $\alpha = 1$, $\beta = 6$, and $\gamma = -48$.

Step 9: Calculate $\alpha + \beta - \gamma$.

$\alpha + \beta - \gamma = 1 + 6 - (-48) = 1 + 6 + 48 = 55$

Common Mistakes & Tips

• Be careful with signs when applying integration by parts.
• Remember to simplify the expression for $g'(x)$ before substituting $x = \frac{\pi}{2}$.
• Double-check your calculations to avoid arithmetic errors.

Summary

We used integration by parts repeatedly to find the indefinite integral of $x^3 \sin x$. We then identified $g(x)$ and calculated $g(\frac{\pi}{2})$ and $g'(\frac{\pi}{2})$. Finally, we found $\alpha, \beta, \gamma$ and calculated $\alpha + \beta - \gamma$, which equals 55.

Final Answer

The final answer is \boxed{55}, which corresponds to option (B).