Logo
Jeeify
NVAEasyJEE 2026Definite Integrals

JEE Mathematics 2026 Question with Solution

6∫0π(sin⁡3x+sin⁡2x+sin⁡x)dx6 \int_{0}^{\pi} (\sin 3x + \sin 2x + \sin x)dx is equal to:

Answer

Correct answer:16

Step-by-step solution

Standard Method

Given:

I=∫0π(sin⁡3x+sin⁡2x+sin⁡x) dxI = \int_{0}^{\pi} (\sin 3x + \sin 2x + \sin x) \, dx

We need to find 6I6I.

Find: The value of 6∫0π(sin⁡3x+sin⁡2x+sin⁡x) dx6 \int_{0}^{\pi} (\sin 3x + \sin 2x + \sin x) \, dx.

Using

∫sin⁡(ax) dx=−1acos⁡(ax)+C\int \sin(ax) \, dx = -\frac{1}{a}\cos(ax) + C

we get

I=[−cos⁡(3x)3−cos⁡(2x)2−cos⁡x]0πI = \left[-\frac{\cos(3x)}{3} - \frac{\cos(2x)}{2} - \cos x\right]_{0}^{\pi}

At x=πx = \pi,

−cos⁡(3π)3−cos⁡(2π)2−cos⁡(π)=−−13−12−(−1)=13−12+1-\frac{\cos(3\pi)}{3} - \frac{\cos(2\pi)}{2} - \cos(\pi) = -\frac{-1}{3} - \frac{1}{2} - (-1) = \frac{1}{3} - \frac{1}{2} + 1

At x=0x = 0,

−cos⁡03−cos⁡02−cos⁡0=−13−12−1-\frac{\cos 0}{3} - \frac{\cos 0}{2} - \cos 0 = -\frac{1}{3} - \frac{1}{2} - 1

Therefore,

I=(13−12+1)−(−13−12−1)=13−12+1+13+12+1=23+2=83\begin{aligned} I &= \left(\frac{1}{3} - \frac{1}{2} + 1\right) - \left(-\frac{1}{3} - \frac{1}{2} - 1\right) \\ &= \frac{1}{3} - \frac{1}{2} + 1 + \frac{1}{3} + \frac{1}{2} + 1 \\ &= \frac{2}{3} + 2 = \frac{8}{3} \end{aligned}

Hence,

6I=6×83=166I = 6 \times \frac{8}{3} = 16

Therefore, the value of the expression is 1616. The solution lists answer key as 4040, but the worked solution evaluates the integral to 1616, so 1616 is taken as the answer.

Property of Sine Integrals

Given:

6∫0π(sin⁡3x+sin⁡2x+sin⁡x) dx6 \int_{0}^{\pi} (\sin 3x + \sin 2x + \sin x) \, dx

Find: Its numerical value.

Use the property over [0,π][0,\pi]:

  • ∫0πsin⁡(nx) dx=0\int_{0}^{\pi} \sin(nx) \, dx = 0 for even nn
  • ∫0πsin⁡(nx) dx=2n\int_{0}^{\pi} \sin(nx) \, dx = \frac{2}{n} for odd nn

So,

∫0πsin⁡3x dx=23,∫0πsin⁡2x dx=0,∫0πsin⁡x dx=2\int_{0}^{\pi} \sin 3x \, dx = \frac{2}{3}, \qquad \int_{0}^{\pi} \sin 2x \, dx = 0, \qquad \int_{0}^{\pi} \sin x \, dx = 2

Thus,

∫0π(sin⁡3x+sin⁡2x+sin⁡x) dx=23+0+2=83\int_{0}^{\pi} (\sin 3x + \sin 2x + \sin x) \, dx = \frac{2}{3} + 0 + 2 = \frac{8}{3}

Therefore,

6×83=166 \times \frac{8}{3} = 16

So the required value is 1616.

Common mistakes

  • Using the raw listed answer 4040 without checking the working. This is incorrect because the solution evaluation gives 83\frac{8}{3} for the integral and hence 1616 after multiplying by 66. Always trust consistent working over a mismatched answer label.

  • Assuming ∫0πsin⁡(2x) dx=22=1\int_{0}^{\pi} \sin(2x) \, dx = \frac{2}{2} = 1. This is wrong because for even nn, ∫0πsin⁡(nx) dx=0\int_{0}^{\pi} \sin(nx) \, dx = 0. Check parity before applying the shortcut property.

  • Forgetting to multiply the integral value by 66 after computing I=∫0π(sin⁡3x+sin⁡2x+sin⁡x) dxI = \int_{0}^{\pi} (\sin 3x + \sin 2x + \sin x) \, dx. The integral is 83\frac{8}{3}, but the question asks for 6I6I, not just II.

Practice more Definite Integrals questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trialMore on Definite Integrals

Related questions