Let $$\int_{}^{} x^{3}\sin x dx =g(x)+C$$, where is the constant of integration. If $$8(g(\frac{\pi}{2})+g'(\frac{\pi}{2}))=\alpha \pi^{3} + \beta \pi^{2} + \gamma ,\alpha ,\beta ,\gamma \in Z$$, then $$\alpha +\beta - \gamma$$ equals :

By the Fundamental Theorem of Calculus, $$g'(x) = x^3 \sin x$$.

To find $$g(x)$$, we use integration by parts repeatedly:

$$\int x^3 \sin x \, dx = -x^3 \cos x + 3x^2 \sin x + 6x \cos x - 6 \sin x + C$$

So, $$g(x) = -x^3 \cos x + 3x^2 \sin x + 6x \cos x - 6 \sin x$$.

2. Evaluate at $$x = \frac{\pi}{2}$$

• $$g'(\frac{\pi}{2}) = (\frac{\pi}{2})^3 \sin(\frac{\pi}{2}) = \frac{\pi^3}{8}(1) = \frac{\pi^3}{8}$$

• $$g(\frac{\pi}{2}) = -(\frac{\pi}{2})^3(0) + 3(\frac{\pi}{2})^2(1) + 6(\frac{\pi}{2})(0) - 6(1) = \frac{3\pi^2}{4} - 6$$

3. Solve for Constants

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

Comparing this to $$\alpha\pi^3 + \beta\pi^2 + \gamma$$, we get:

$$\alpha = 1, \beta = 6, \gamma = -48$$.

Final Calculation: $$\alpha + \beta - \gamma = 1 + 6 - (-48) = 55$$.