Indefinite Integration by parts – JEE Main – 23 January 2025 Shift 2

Solution:

Let $I = \int x^3 \sin x dx$. We use integration by parts $\int u dv = uv – \int v du$.

First iteration:
Let $u = x^3 \Rightarrow du = 3x^2 dx$
Let $dv = \sin x dx \Rightarrow v = -\cos x$
$$I = -x^3 \cos x – \int (-\cos x)(3x^2) dx = -x^3 \cos x + 3 \int x^2 \cos x dx$$

Second iteration (for $\int x^2 \cos x dx$):
Let $u = x^2 \Rightarrow du = 2x dx$
Let $dv = \cos x dx \Rightarrow v = \sin x$
$$\int x^2 \cos x dx = x^2 \sin x – \int 2x \sin x dx$$
Substitute back:
$$I = -x^3 \cos x + 3(x^2 \sin x – 2 \int x \sin x dx)$$
$$I = -x^3 \cos x + 3x^2 \sin x – 6 \int x \sin x dx$$

Third iteration (for $\int x \sin x dx$):
Let $u = x \Rightarrow du = dx$
Let $dv = \sin x dx \Rightarrow v = -\cos x$
$$\int x \sin x dx = -x \cos x – \int (-\cos x) dx = -x \cos x + \sin x$$
Substitute back:
$$I = -x^3 \cos x + 3x^2 \sin x – 6(-x \cos x + \sin x) + C$$
$$g(x) = -x^3 \cos x + 3x^2 \sin x + 6x \cos x – 6 \sin x$$

Now we calculate $g(\frac{\pi}{2})$ and $g'(\frac{\pi}{2})$.
$$g\left(\frac{\pi}{2}\right) = -\left(\frac{\pi}{2}\right)^3 \cos\frac{\pi}{2} + 3\left(\frac{\pi}{2}\right)^2 \sin\frac{\pi}{2} + 6\left(\frac{\pi}{2}\right) \cos\frac{\pi}{2} – 6 \sin\frac{\pi}{2}$$
Since $\cos\frac{\pi}{2} = 0$ and $\sin\frac{\pi}{2} = 1$:
$$g\left(\frac{\pi}{2}\right) = 0 + 3\left(\frac{\pi^2}{4}\right)(1) + 0 – 6(1) = \frac{3\pi^2}{4} – 6$$

For $g'(x)$, we know that $g(x) = \int x^3 \sin x dx$, so by fundamental theorem of calculus:
$$g'(x) = x^3 \sin x$$
$$g’\left(\frac{\pi}{2}\right) = \left(\frac{\pi}{2}\right)^3 \sin\frac{\pi}{2} = \frac{\pi^3}{8}$$

Substitute into the given expression:
$$8\left(g\left(\frac{\pi}{2}\right) + g’\left(\frac{\pi}{2}\right)\right) = 8\left(\left(\frac{3\pi^2}{4} – 6\right) + \frac{\pi^3}{8}\right)$$
$$= 8\left(\frac{3\pi^2}{4}\right) – 8(6) + 8\left(\frac{\pi^3}{8}\right)$$
$$= 6\pi^2 – 48 + \pi^3$$
$$= \pi^3 + 6\pi^2 – 48$$

Comparing with $\alpha\pi^3 + \beta\pi^2 + \gamma$:
$\alpha = 1, \beta = 6, \gamma = -48$.

We need to find $\alpha + \beta – \gamma$:
$$\alpha + \beta – \gamma = 1 + 6 – (-48) = 7 + 48 = 55$$

Ans. (55)