JEE MainMathematics

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+4 marks−1 if incorrectSingle correctPrevious-year question

Integration by Parts of $x^{n} \sin x$

Q: 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.

Let integral of x^3 sin x dx = g(x) + C, where C is the constant of integration. If 8 (g (π/2) + g' (π/2)) = απ^3 + βπ^2 + γ, α, β, γ ∈ ℤ, then α + β - γ equals:

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A48B55C62D47

Step-by-step solutionView

Correct answer

Integrating $x^{3} \sin x$ by parts yields $g (x) = - x^{3} \cos x + 3 x^{2} \sin x + 6 x \cos x - 6 \sin x$ , from which $8 (g (\frac{π}{2}) + g^{'} (\frac{π}{2})) = π^{3} + 6 π^{2} - 48$ , giving $α = 1, β = 6, γ = - 48$ , and $α + β - γ = 55$ .

Option analysis

Why each option works or fails

A · 48

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

B · 55

This is the correct calculation. Correctly compute $g (x)$ via repeated integration by parts, evaluate $g (π / 2) + g^{'} (π / 2)$ , match coefficients with the polynomial in $π$ , and evaluate $α + β - γ$ .

C · 62

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

D · 47

Taking $α + β + γ = 1 + 6 - 48 = - 41$ or subtracting $α + β$ from $- γ$ incorrectly to get $48 - 1 = 47$ . Read the exact expression $α + β - γ = 1 + 6 - (- 48) = 7 + 48 = 55$ .

Reviewed route

Solution

StepWorking

01Given

Given the indefinite integral $\int x^{3} \sin x d x = g (x) + C$ , and the condition $8 (g (\frac{π}{2}) + g^{'} (\frac{π}{2})) = α π^{3} + β π^{2} + γ$ with $α, β, γ \in Z$ .

02Goal

Find the values of $α, β, γ$ and compute $α + β - γ$ .

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 $x = \frac{π}{2}$ and equate coefficients.

04Execute

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

05Execute

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

06Execute

Multiply the sum by 8: $8 (g (\frac{π}{2}) + g^{'} (\frac{π}{2})) = 8 (\frac{π^{3}}{8} + \frac{3 π^{2}}{4} - 6) = π^{3} + 6 π^{2} - 48$ Comparing with $α π^{3} + β π^{2} + γ$ , we find: $α = 1, β = 6, γ = - 48$ Thus: $α + β - γ = 1 + 6 - (- 48) = 7 + 48 = 55$

✓Verify

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

Hints that build this answer step by step

What is the antiderivative $g (x) = \int x^{3} \sin x d x$ ?
$g (x) = - x^{3} \cos x + 3 x^{2} \sin x + 6 x \cos x - 6 \sin x$
What are the values of $g (\frac{π}{2})$ and $g^{'} (\frac{π}{2})$ ?
$g (\frac{π}{2}) = \frac{3 π^{2}}{4} - 6$ and $g^{'} (\frac{π}{2}) = \frac{π^{3}}{8}$
Given $8 (g (\frac{π}{2}) + g^{'} (\frac{π}{2})) = π^{3} + 6 π^{2} - 48 = α π^{3} + β π^{2} + γ$ , what is $α + β - γ$ ?
$α + β - γ = 1 + 6 - (- 48) = 55$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
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 $g (x) = - x^{3} \cos x + 3 x^{2} \sin x + 6 x \cos x - 6 \sin x$ , from which $8 (g (\frac{π}{2}) + g^{'} (\frac{π}{2})) = π^{3} + 6 π^{2} - 48$ , giving $α = 1, β = 6, γ = - 48$ , and $α + β - γ = 55$ .

Why each option works or fails

A: 48 - Solving for $γ$ and finding $- γ = 48$ , but forgetting the remaining terms $α + β$ , resulting in 48. Ensure you compute the complete requested linear combination $α + β - γ$ rather than just reading off $- γ$ .
B · correct: 55 - This is the correct calculation. Correctly compute $g (x)$ via repeated integration by parts, evaluate $g (π / 2) + g^{'} (π / 2)$ , match coefficients with the polynomial in $π$ , and evaluate $α + β - γ$ .
C: 62 - Making a sign error on $γ$ , taking $γ = 48$ instead of $γ = - 48$ , leading to $1 + 6 - (- 48) \to 1 + 6 + 55$ or a miscalculation like $1 + 6 - 48 + 103$ . Carefully track the negative sign on $- 6 \sin (π / 2) = - 6$ , multiplied by $8$ to give $γ = - 48$ .
D: 47 - Taking $α + β + γ = 1 + 6 - 48 = - 41$ or subtracting $α + β$ from $- γ$ incorrectly to get $48 - 1 = 47$ . Read the exact expression $α + β - γ = 1 + 6 - (- 48) = 7 + 48 = 55$ .

Step-by-step solution

Given: Given the indefinite integral $\int x^{3} \sin x d x = g (x) + C$ , and the condition $8 (g (\frac{π}{2}) + g^{'} (\frac{π}{2})) = α π^{3} + β π^{2} + γ$ with $α, β, γ \in Z$ .
Goal: Find the values of $α, β, γ$ and compute $α + β - γ$ .
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 $x = \frac{π}{2}$ and equate coefficients.
Execute: Applying tabular integration by parts to $\int x^{3} \sin x d x$ : $g (x) = - x^{3} \cos x + 3 x^{2} \sin x + 6 x \cos x - 6 \sin x$ Also, from the definition of an antiderivative: $g^{'} (x) = x^{3} \sin x$
Execute: Evaluate $g (\frac{π}{2})$ and $g^{'} (\frac{π}{2})$ using $\cos (\frac{π}{2}) = 0$ and $\sin (\frac{π}{2}) = 1$ : $g (\frac{π}{2}) = 0 + 3 {(\frac{π}{2})}^{2} (1) + 0 - 6 (1) = \frac{3 π^{2}}{4} - 6$ $g^{'} (\frac{π}{2}) = {(\frac{π}{2})}^{3} (1) = \frac{π^{3}}{8}$
Execute: Multiply the sum by 8: $8 (g (\frac{π}{2}) + g^{'} (\frac{π}{2})) = 8 (\frac{π^{3}}{8} + \frac{3 π^{2}}{4} - 6) = π^{3} + 6 π^{2} - 48$ Comparing with $α π^{3} + β π^{2} + γ$ , we find: $α = 1, β = 6, γ = - 48$ Thus: $α + β - γ = 1 + 6 - (- 48) = 7 + 48 = 55$
Verify: Differentiating $g (x) = - x^{3} \cos x + 3 x^{2} \sin x + 6 x \cos x - 6 \sin x$ gives $g^{'} (x) = (- 3 x^{2} \cos x + x^{3} \sin x) + (6 x \sin x + 3 x^{2} \cos x) + (6 \cos x - 6 x \sin x) - 6 \cos x = x^{3} \sin x$ , which is verified.

Integration by Parts of $x^{n} \sin x$

What feels right?

Integrating $x^{3} \sin x$ by parts yields $g (x) = - x^{3} \cos x + 3 x^{2} \sin x + 6 x \cos x - 6 \sin x$ , from which $8 (g (\frac{π}{2}) + g^{'} (\frac{π}{2})) = π^{3} + 6 π^{2} - 48$ , giving $α = 1, β = 6, γ = - 48$ , and $α + β - γ = 55$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

06Execute

✓Verify

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Integration by Parts of xnsin⁡xxnsinx

Integrating x3sin⁡xx3sinx by parts yields g(x)=−x3cos⁡x+3x2sin⁡x+6xcos⁡x−6sin⁡xg(x)=−x3cosx+3x2sinx+6xcosx−6sinx, from which 8(g(π2)+g′(π2))=π3+6π2−488(g(2π​)+g′(2π​))=π3+6π2−48, giving α=1,β=6,γ=−48α=1,β=6,γ=−48, and α+β−γ=55α+β−γ=55.

01Given

02Goal

03Approach

04Execute

05Execute

06Execute

✓Verify

The value of the integral…

The same chapter, from another paper — Integral Calculus, 2025

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Students also ask

Integration by Parts of $x^{n} \sin x$

Integrating $x^{3} \sin x$ by parts yields $g (x) = - x^{3} \cos x + 3 x^{2} \sin x + 6 x \cos x - 6 \sin x$ , from which $8 (g (\frac{π}{2}) + g^{'} (\frac{π}{2})) = π^{3} + 6 π^{2} - 48$ , giving $α = 1, β = 6, γ = - 48$ , and $α + β - γ = 55$ .