Indefinite Integration
Indefinite Integration
nta_pyq_2025_apr
Grade 12
Question:
Let $\displaystyle\int x^3\sin x\,dx = g(x)+C$, where $C$ is the constant of integration. If $8\!\left(g\!\left(\dfrac{\pi}{2}\right)+g'\!\left(\dfrac{\pi}{2}\right)\right) = \alpha\pi^3+\beta\pi^2+\gamma$, $\alpha,\beta,\gamma\in\mathbb{Z}$, then $\alpha+\beta-\gamma$ equals:
Step-by-Step Solution
Key Concept: Apply integration by parts three times to get $g(x) = -x^3\cos x + 3x^2\sin x + 6x\cos x - 6\sin x$, then note that $g'(x) = x^3\sin x$ (the original integrand), so $g'(\pi/2) = (\pi/2)^3$.
Using IBP repeatedly:
$$g(x) = -x^3\cos x + 3x^2\sin x + 6x\cos x - 6\sin x.$$
$$g\!\left(\frac{\pi}{2}\right) = 0 + \frac{3\pi^2}{4} + 0 - 6 = \frac{3\pi^2}{4}-6.$$
$$g'(x) = x^3\sin x \Rightarrow g'\!\left(\frac{\pi}{2}\right) = \frac{\pi^3}{8}.$$
$$8\!\left(g\!\left(\frac{\pi}{2}\right)+g'\!\left(\frac{\pi}{2}\right)\right) = 8\!\left(\frac{3\pi^2}{4}-6+\frac{\pi^3}{8}\right) = \pi^3+6\pi^2-48.$$
So $\alpha=1$, $\beta=6$, $\gamma=-48$. Hence $\alpha+\beta-\gamma = 1+6+48 = 55$.
Correct Answer: 2