JEE Main 2025 Jan 23 Shift 2, Mathematics Q4: Integration by Parts

JEE Main2025Jan 23, Shift 2Mathematics

Let $\int x^{3} sin x d x = g (x) + C$ , where $C$ is the constant of integration.

If $8 (g (\frac{π}{2}) + g^{'} (\frac{π}{2})) = α π^{3} + β π^{2} + γ$ , where $α, β, γ \in Z$ , then $α + β - γ$ equals :

Solution

Integrate by parts repeatedly (or use tabular integration). The antiderivative simplifies to

$g (x) = - x^{3} cos x + 3 x^{2} sin x + 6 x cos x - 6 sin x .$

Differentiating gives back $g^{'} (x) = x^{3} sin x$ (check: all middle terms cancel).

Evaluate at $x = \frac{π}{2}$ (where $cos x = 0$ , $sin x = 1$ ):

$g (\frac{π}{2}) = 0 + 3 \cdot \frac{π ^{2}}{4} \cdot 1 + 0 - 6 = \frac{3 π ^{2}}{4} - 6.$

$g^{'} (\frac{π}{2}) = (\frac{π}{2})^{3} \cdot 1 = \frac{π ^{3}}{8} .$

$8 (g (\frac{π}{2}) + g^{'} (\frac{π}{2})) = 8 (\frac{3 π ^{2}}{4} - 6 + \frac{π ^{3}}{8}) = π^{3} + 6 π^{2} - 48.$

So $α = 1, β = 6, γ = - 48$ , giving $α + β - γ = 1 + 6 + 48 = 55$ .

Concept behind this question

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