Let int x^3 sin x , dx = g(x) + C,where C is | vedclass

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(A) Using integration by parts,$\int u \, dv = uv - \int v \, du$. Let $u = x^3$ and $dv = \sin x \, dx$. Then $du = 3x^2 \, dx$ and $v = -\cos x$.$\i

Vedclass · Accepted Answer

$55$

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(A) Using integration by parts,$\int u \, dv = uv - \int v \, du$. Let $u = x^3$ and $dv = \sin x \, dx$. Then $du = 3x^2 \, dx$ and $v = -\cos x$.$\int x^3 \sin x \, dx = -x^3 \cos x + \int 3x^2 \cos x \, dx$.Applying integration by parts again for $\int 3x^2 \cos x \, dx$: $u = 3x^2, dv = \cos x \, dx \implies du = 6x \, dx, v = \sin x$.$= -x^3 \cos x + 3x^2 \sin x - \int 6x \sin x \, dx$.Applying integration by parts for $\int 6x \sin x \, dx$: $u = 6x, dv = \sin x \, dx \implies du = 6 \, dx, v = -\cos x$.$= -x^3 \cos x + 3x^2 \sin x - (6x(-\cos x) - \int -6 \cos x \, dx) = -x^3 \cos x + 3x^2 \sin x + 6x \cos x - 6 \sin x + C$.Thus,$g(x) = -x^3 \cos x + 3x^2 \sin x + 6x \cos x - 6 \sin x$.$g\left(\frac{\pi}{2}\right) = -\left(\frac{\pi}{2}\right)^3 \cos\left(\frac{\pi}{2}\right) + 3\left(\frac{\pi}{2}\right)^2 \sin\left(\frac{\pi}{2}\right) + 6\left(\frac{\pi}{2}\right) \cos\left(\frac{\pi}{2}\right) - 6 \sin\left(\frac{\pi}{2}\right) = 0 + \frac{3\pi^2}{4} + 0 - 6 = \frac{3\pi^2}{4} - 6$.Since $g(x) = \int x^3 \sin x \, dx$,by the Fundamental Theorem of Calculus,$g^{\prime}(x) = x^3 \sin x$.$g^{\prime}\left(\frac{\pi}{2}\right) = \left(\frac{\pi}{2}\right)^3 \sin\left(\frac{\pi}{2}\right) = \frac{\pi^3}{8}$.$8\left(g\left(\frac{\pi}{2}\right) + g^{\prime}\left(\frac{\pi}{2}\right)\right) = 8\left(\frac{3\pi^2}{4} - 6 + \frac{\pi^3}{8}\right) = 6\pi^2 - 48 + \pi^3 = 1\pi^3 + 6\pi^2 - 48$.Comparing with $\alpha \pi^3 + \beta \pi^2 + \gamma$,we get $\alpha = 1, \beta = 6, \gamma = -48$.$\alpha + \beta - \gamma = 1 + 6 - (-48) = 7 + 48 = 55$.

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

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