JEE MainMathematics

Answer verified against the official NTA key

+4 marks−1 if incorrectNumericalPrevious-year question

Definite Integration of Trigonometric Absolute Value Function

Q: Why can we remove |sin x| as just sin x?

For x ∈ [0, π], sin x ≥ 0, so |sin x| = sin x everywhere in the integration interval.

6 integral from 0 to π of |(sin 3x + sin 2x + sin x)| dx is equal to.........

Your answer stays private

What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

Type the value - units or words beside it are fine.

Step-by-step solutionView

Correct answer

The value of $6 \int_{0}^{π} ∣ \sin 3 x + \sin 2 x + \sin x ∣ d x$ is 17.

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01Given

The expression to evaluate is $I = 6 \int_{0}^{π} ∣ \sin 3 x + \sin 2 x + \sin x ∣ d x$ .

02Approach

Combine $\sin 3 x + \sin x = 2 \sin 2 x \cos x$ to factor out $\sin 2 x$ , then express entirely in terms of $\sin x$ and $\cos x$ . Since $\sin x \geq 0$ on $[0, π]$ , factor out $\sin x$ from the absolute value and use the substitution $t = \cos x$ .

03Execute

Factor the integrand: $\sin 3 x + \sin x + \sin 2 x = 2 \sin 2 x \cos x + \sin 2 x = \sin 2 x (2 \cos x + 1) = 2 \sin x \cos x (2 \cos x + 1) = 2 \sin x (2 \cos^{2} x + \cos x)$ . Thus, $I = 12 \int_{0}^{π} \sin x ∣ 2 \cos^{2} x + \cos x ∣ d x$ .

04Execute

Substitute $t = \cos x$ , so $d t = - \sin x d x$ . When $x = 0$ , $t = 1$ ; when $x = π$ , $t = - 1$ . The integral becomes $I = 12 \int_{- 1}^{1} ∣ 2 t^{2} + t ∣ d t$ .

05Execute

Analyze the sign of $2 t^{2} + t = t (2 t + 1)$ : it is positive on $[- 1, - 1 / 2]$ and $[0, 1]$ , and negative on $(- 1 / 2, 0)$ . Split the integral: $I = 12 [\int_{- 1}^{- 1 / 2} (2 t^{2} + t) d t - \int_{- 1 / 2}^{0} (2 t^{2} + t) d t + \int_{0}^{1} (2 t^{2} + t) d t]$ . Computing the antiderivative $F (t) = \frac{2}{3} t^{3} + \frac{1}{2} t^{2}$ : $F (- 1) = - 1 / 6$ , $F (- 1 / 2) = 1 / 24$ , $F (0) = 0$ , $F (1) = 7 / 6$ . Therefore, $I = 12 [(1 / 24 - (- 1 / 6)) - (0 - 1 / 24) + (7 / 6 - 0)] = 12 [5 / 24 + 1 / 24 + 28 / 24] = 12 \times \frac{34}{24} = 17$ .

✓Verify

Check piecewise: $\int_{- 1}^{- 1 / 2} ∣ 2 t^{2} + t ∣ d t = 5 / 24$ , $\int_{- 1 / 2}^{0} ∣ 2 t^{2} + t ∣ d t = 1 / 24$ , $\int_{0}^{1} ∣ 2 t^{2} + t ∣ d t = 7 / 6 = 28 / 24$ . Sum = $34 / 24$ . Multiplying by 12 gives exactly 17.

Hints that build this answer step by step

How should the integrand $\sin 3 x + \sin 2 x + \sin x$ be factored to determine where its sign changes on $[0, π]$ ?
Combine $\sin 3 x + \sin x = 2 \sin 2 x \cos x$ to factor out $\sin 2 x$ , giving $\sin 2 x (2 \cos x + 1)$ .
At which points in $(0, π)$ does the integrand $2 \sin x \cos x (2 \cos x + 1)$ change sign?
$x = \frac{π}{2}$ and $x = \frac{2 π}{3}$
What is the antiderivative $F (x) = \int (\sin 3 x + \sin 2 x + \sin x) d x$ and the sign in each interval?
$F (x) = - \frac{\cos 3 x}{3} - \frac{\cos 2 x}{2} - \cos x$ , positive on $(0, \frac{π}{2})$ and $(\frac{2 π}{3}, π)$ , negative on $(\frac{π}{2}, \frac{2 π}{3})$ .
Evaluating the split integral and multiplying by 6 gives what final value?
17

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 19 September 2026

Quick checks

Students also ask

Why can we remove

∣ \sin x ∣

as just

\sin x

For $x \in [0, π]$ , $\sin x \geq 0$ , so $∣ \sin x ∣ = \sin x$ everywhere in the integration interval.

Answer

Accepted answer: 17

The value of $6 \int_{0}^{π} ∣ \sin 3 x + \sin 2 x + \sin x ∣ d x$ is 17.

Why each option works or fails

Step-by-step solution

Given: The expression to evaluate is $I = 6 \int_{0}^{π} ∣ \sin 3 x + \sin 2 x + \sin x ∣ d x$ .
Approach: Combine $\sin 3 x + \sin x = 2 \sin 2 x \cos x$ to factor out $\sin 2 x$ , then express entirely in terms of $\sin x$ and $\cos x$ . Since $\sin x \geq 0$ on $[0, π]$ , factor out $\sin x$ from the absolute value and use the substitution $t = \cos x$ .
Execute: Factor the integrand: $\sin 3 x + \sin x + \sin 2 x = 2 \sin 2 x \cos x + \sin 2 x = \sin 2 x (2 \cos x + 1) = 2 \sin x \cos x (2 \cos x + 1) = 2 \sin x (2 \cos^{2} x + \cos x)$ . Thus, $I = 12 \int_{0}^{π} \sin x ∣ 2 \cos^{2} x + \cos x ∣ d x$ .
Execute: Substitute $t = \cos x$ , so $d t = - \sin x d x$ . When $x = 0$ , $t = 1$ ; when $x = π$ , $t = - 1$ . The integral becomes $I = 12 \int_{- 1}^{1} ∣ 2 t^{2} + t ∣ d t$ .
Execute: Analyze the sign of $2 t^{2} + t = t (2 t + 1)$ : it is positive on $[- 1, - 1 / 2]$ and $[0, 1]$ , and negative on $(- 1 / 2, 0)$ . Split the integral: $I = 12 [\int_{- 1}^{- 1 / 2} (2 t^{2} + t) d t - \int_{- 1 / 2}^{0} (2 t^{2} + t) d t + \int_{0}^{1} (2 t^{2} + t) d t]$ . Computing the antiderivative $F (t) = \frac{2}{3} t^{3} + \frac{1}{2} t^{2}$ : $F (- 1) = - 1 / 6$ , $F (- 1 / 2) = 1 / 24$ , $F (0) = 0$ , $F (1) = 7 / 6$ . Therefore, $I = 12 [(1 / 24 - (- 1 / 6)) - (0 - 1 / 24) + (7 / 6 - 0)] = 12 [5 / 24 + 1 / 24 + 28 / 24] = 12 \times \frac{34}{24} = 17$ .
Verify: Check piecewise: $\int_{- 1}^{- 1 / 2} ∣ 2 t^{2} + t ∣ d t = 5 / 24$ , $\int_{- 1 / 2}^{0} ∣ 2 t^{2} + t ∣ d t = 1 / 24$ , $\int_{0}^{1} ∣ 2 t^{2} + t ∣ d t = 7 / 6 = 28 / 24$ . Sum = $34 / 24$ . Multiplying by 12 gives exactly 17.

Definite Integration of Trigonometric Absolute Value Function

What feels right?

The value of $6 \int_{0}^{π} ∣ \sin 3 x + \sin 2 x + \sin x ∣ d x$ is 17.

Why each option works or fails

Solution

01Given

02Approach

03Execute

04Execute

05Execute

✓Verify

We think you should solve this next.

The same chapter, from another paper — Integral Calculus, 2023

The area of the region given by ${(x, y) : x y \leq 8, 1 \leq y \leq x^{2}}$ is…

This chapter, in a different year — Integral Calculus, 2023

Students also ask

Definite Integration of Trigonometric Absolute Value Function

The value of 6∫0π∣sin⁡3x+sin⁡2x+sin⁡x∣dx6∫0π​∣sin3x+sin2x+sinx∣dx is 17.

01Given

02Approach

03Execute

04Execute

05Execute

✓Verify

The same chapter, from another paper — Integral Calculus, 2023

The area of the region given by {(x,y):xy≤8,1≤y≤x2}{(x,y):xy≤8,1≤y≤x2} is…

This chapter, in a different year — Integral Calculus, 2023

Students also ask

The value of $6 \int_{0}^{π} ∣ \sin 3 x + \sin 2 x + \sin x ∣ d x$ is 17.

The area of the region given by ${(x, y) : x y \leq 8, 1 \leq y \leq x^{2}}$ is…