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JEE Main 202621 JanuaryMorning Shift
Exam
JEE Main
Year
2026
Date
21 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Definite Integration

JEE Main 2026 · 21 January · Morning Shift · Mathematics · Definite Integration

6 ∫_ 0 ^ π |( 3 x+ 2 x+ x)| d x is equal to \_\_\_\_ .

6∫0π∣(sin⁡3x+sin⁡2x+sin⁡x)∣dx6 \int_{0}^{\pi}|(\sin 3 x+\sin 2 x+\sin x)| d x is equal to ____\_\_\_\_.

Correct answer

0

Step-by-step solution

Using sin⁡3x+sin⁡x=2sin⁡2xcos⁡x\sin 3x + \sin x = 2\sin 2x \cos x:
sin⁡3x+sin⁡2x+sin⁡x=sin⁡2x(2cos⁡x+1)=2sin⁡xcos⁡x(2cos⁡x+1)\sin 3x + \sin 2x + \sin x = \sin 2x(2\cos x + 1) = 2\sin x \cos x(2\cos x + 1)
Zeros in [0,π][0, \pi]: x=0,π/2,2π/3,πx = 0, \pi/2, 2\pi/3, \pi
Expression is positive on (0,π/2)(0, \pi/2), negative on (π/2,2π/3)(\pi/2, 2\pi/3), positive on (2π/3,π)(2\pi/3, \pi)
Let F(x)=−cos⁡3x3−cos⁡2x2−cos⁡xF(x) = -\frac{\cos 3x}{3} - \frac{\cos 2x}{2} - \cos x
F(0)=−116F(0) = -\frac{11}{6}, F(π/2)=12F(\pi/2) = \frac{1}{2}, F(2π/3)=512F(2\pi/3) = \frac{5}{12}, F(π)=56F(\pi) = \frac{5}{6}
∫0π/2=73\int_0^{\pi/2} = \frac{7}{3}, \left|\int_{\pi/2}^{2\pi/3}\right| = \frac{1}{12},,\int_{2\pi/3}^{\pi} = \frac{5}{12}<br>Total<br>Total= \frac{7}{3} + \frac{1}{12} + \frac{5}{12} = \frac{17}{6}<br><br>6 \times \frac{17}{6} = 17
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