Do we need to explicitly compute the derivative using the formula of ?

No, because is an antiderivative of , so by the Fundamental Theorem of Calculus, directly.

JEE MainMathematics

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Integral Calculus: JEE Main Mathematics Question with Solution

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' \left(\frac{\pi}{2}\right)\right) = \alpha\pi^{3} + \beta\pi^{2} + \gamma

\alpha, \beta, \gamma \in \mathbb{Z}

, then

\alpha + \beta - \gamma

equals :

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A48B55C62D47

Step-by-step solutionView

Correct answer

Integrating $x^3 \sin x$ by parts yields $g(x) = -x^3 \cos x + 3x^2 \sin x + 6x \cos x - 6 \sin x$ , from which $8\left(g\left(\frac{\pi}{2}\right) + g'\left(\frac{\pi}{2}\right)\right) = \pi^3 + 6\pi^2 - 48$ , giving $\alpha=1, \beta=6, \gamma=-48$ , and $\alpha + \beta - \gamma = 55$ .

Option analysis

Why each option works or fails

A · 48

Solving for $\gamma$ and finding $-\gamma = 48$ , but forgetting the remaining terms $\alpha + \beta$ , resulting in 48. Ensure you compute the complete requested linear combination $\alpha + \beta - \gamma$ rather than just reading off $-\gamma$ .

B · 55

None. This is the correct calculation. Correctly compute $g(x)$ via repeated integration by parts, evaluate $g(\pi/2) + g'(\pi/2)$ , match coefficients with the polynomial in $\pi$ , and evaluate $\alpha + \beta - \gamma$ .

C · 62

Making a sign error on $\gamma$ , taking $\gamma = 48$ instead of $\gamma = -48$ , leading to $1 + 6 - (-48) \rightarrow 1 + 6 + 55$ or a miscalculation like $1 + 6 - 48 + 103$ . Carefully track the negative sign on $-6\sin(\pi/2) = -6$ , multiplied by $8$ to give $\gamma = -48$ .

D · 47

Taking $\alpha + \beta + \gamma = 1 + 6 - 48 = -41$ or subtracting $\alpha + \beta$ from $-\gamma$ incorrectly to get $48 - 1 = 47$ . Read the exact expression $\alpha + \beta - \gamma = 1 + 6 - (-48) = 7 + 48 = 55$ .

Reviewed route

Solution

StepWorking

01given

Given the indefinite integral $\int x^3 \sin x \, dx = g(x) + C$ , and the condition $8 \left(g \left(\frac{\pi}{2}\right) + g' \left(\frac{\pi}{2}\right)\right) = \alpha\pi^3 + \beta\pi^2 + \gamma$ with $\alpha, \beta, \gamma \in \mathbb{Z}$ .

02goal

Find the values of $\alpha, \beta, \gamma$ and compute $\alpha + \beta - \gamma$ .

03approach

Integrate $x^3 \sin x$ using repeated integration by parts (or tabular integration) to determine $g(x)$ . Use the Fundamental Theorem of Calculus to note $g'(x) = x^3 \sin x$ . Then evaluate both at $x = \frac{\pi}{2}$ and equate coefficients.

04execute

Applying tabular integration by parts to $\int x^3 \sin x \, dx$ : $g(x) = -x^3 \cos x + 3x^2 \sin x + 6x \cos x - 6 \sin x$ Also, from the definition of an antiderivative: $g'(x) = x^3 \sin x$

05execute

Evaluate $g\left(\frac{\pi}{2}\right)$ and $g'\left(\frac{\pi}{2}\right)$ using $\cos\left(\frac{\pi}{2}\right) = 0$ and $\sin\left(\frac{\pi}{2}\right) = 1$ : $g\left(\frac{\pi}{2}\right) = 0 + 3\left(\frac{\pi}{2}\right)^2(1) + 0 - 6(1) = \frac{3\pi^2}{4} - 6$ $g'\left(\frac{\pi}{2}\right) = \left(\frac{\pi}{2}\right)^3 (1) = \frac{\pi^3}{8}$

06execute

Multiply the sum by 8: $8\left(g\left(\frac{\pi}{2}\right) + g'\left(\frac{\pi}{2}\right)\right) = 8\left(\frac{\pi^3}{8} + \frac{3\pi^2}{4} - 6\right) = \pi^3 + 6\pi^2 - 48$ Comparing with $\alpha\pi^3 + \beta\pi^2 + \gamma$ , we find: $\alpha = 1, \quad \beta = 6, \quad \gamma = -48$ Thus: $\alpha + \beta - \gamma = 1 + 6 - (-48) = 7 + 48 = 55$

✓verify

Differentiating $g(x) = -x^3\cos x + 3x^2\sin x + 6x\cos x - 6\sin x$ gives $g'(x) = (-3x^2\cos x + x^3\sin x) + (6x\sin x + 3x^2\cos x) + (6\cos x - 6x\sin x) - 6\cos x = x^3\sin x$ , which is verified.

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 7 September 2026

Quick checks

Students also ask

Do we need to explicitly compute the derivative

g'(x)

using the formula of

g(x)

No, because $g(x)$ is an antiderivative of $x^3 \sin x$ , so by the Fundamental Theorem of Calculus, $g'(x) = x^3 \sin x$ directly.

Answer

Integrating $x^3 \sin x$ by parts yields $g(x) = -x^3 \cos x + 3x^2 \sin x + 6x \cos x - 6 \sin x$ , from which $8\left(g\left(\frac{\pi}{2}\right) + g'\left(\frac{\pi}{2}\right)\right) = \pi^3 + 6\pi^2 - 48$ , giving $\alpha=1, \beta=6, \gamma=-48$ , and $\alpha + \beta - \gamma = 55$ .

Why each option works or fails

A: 48 - Solving for $\gamma$ and finding $-\gamma = 48$ , but forgetting the remaining terms $\alpha + \beta$ , resulting in 48. Ensure you compute the complete requested linear combination $\alpha + \beta - \gamma$ rather than just reading off $-\gamma$ .
B · correct: 55 - None. This is the correct calculation. Correctly compute $g(x)$ via repeated integration by parts, evaluate $g(\pi/2) + g'(\pi/2)$ , match coefficients with the polynomial in $\pi$ , and evaluate $\alpha + \beta - \gamma$ .
C: 62 - Making a sign error on $\gamma$ , taking $\gamma = 48$ instead of $\gamma = -48$ , leading to $1 + 6 - (-48) \rightarrow 1 + 6 + 55$ or a miscalculation like $1 + 6 - 48 + 103$ . Carefully track the negative sign on $-6\sin(\pi/2) = -6$ , multiplied by $8$ to give $\gamma = -48$ .
D: 47 - Taking $\alpha + \beta + \gamma = 1 + 6 - 48 = -41$ or subtracting $\alpha + \beta$ from $-\gamma$ incorrectly to get $48 - 1 = 47$ . Read the exact expression $\alpha + \beta - \gamma = 1 + 6 - (-48) = 7 + 48 = 55$ .

Step-by-step solution

given: Given the indefinite integral $\int x^3 \sin x \, dx = g(x) + C$ , and the condition $8 \left(g \left(\frac{\pi}{2}\right) + g' \left(\frac{\pi}{2}\right)\right) = \alpha\pi^3 + \beta\pi^2 + \gamma$ with $\alpha, \beta, \gamma \in \mathbb{Z}$ .
goal: Find the values of $\alpha, \beta, \gamma$ and compute $\alpha + \beta - \gamma$ .
approach: Integrate $x^3 \sin x$ using repeated integration by parts (or tabular integration) to determine $g(x)$ . Use the Fundamental Theorem of Calculus to note $g'(x) = x^3 \sin x$ . Then evaluate both at $x = \frac{\pi}{2}$ and equate coefficients.
execute: Applying tabular integration by parts to $\int x^3 \sin x \, dx$ : $g(x) = -x^3 \cos x + 3x^2 \sin x + 6x \cos x - 6 \sin x$ Also, from the definition of an antiderivative: $g'(x) = x^3 \sin x$
execute: Evaluate $g\left(\frac{\pi}{2}\right)$ and $g'\left(\frac{\pi}{2}\right)$ using $\cos\left(\frac{\pi}{2}\right) = 0$ and $\sin\left(\frac{\pi}{2}\right) = 1$ : $g\left(\frac{\pi}{2}\right) = 0 + 3\left(\frac{\pi}{2}\right)^2(1) + 0 - 6(1) = \frac{3\pi^2}{4} - 6$ $g'\left(\frac{\pi}{2}\right) = \left(\frac{\pi}{2}\right)^3 (1) = \frac{\pi^3}{8}$
execute: Multiply the sum by 8: $8\left(g\left(\frac{\pi}{2}\right) + g'\left(\frac{\pi}{2}\right)\right) = 8\left(\frac{\pi^3}{8} + \frac{3\pi^2}{4} - 6\right) = \pi^3 + 6\pi^2 - 48$ Comparing with $g\left(\frac{\pi}{2}\right)$ , we find: $\alpha = 1, \quad \beta = 6, \quad \gamma = -48$ Thus: $\alpha + \beta - \gamma = 1 + 6 - (-48) = 7 + 48 = 55$
verify: Differentiating $g(x) = -x^3\cos x + 3x^2\sin x + 6x\cos x - 6\sin x$ gives $g'(x) = (-3x^2\cos x + x^3\sin x) + (6x\sin x + 3x^2\cos x) + (6\cos x - 6x\sin x) - 6\cos x = x^3\sin x$ , which is verified.