$6 \int_{0}^{\pi}|(\sin 3 x+\sin 2 x+\sin x)| dx$ is equal to.......

JEE Main · 2026 · 21 Jan 2026 · Shift 1 · Definite Integrals

6 \int_{0}^{π} ∣ (sin 3 x + sin 2 x + sin x) ∣ d x

is equal to.......

Answer: 17

Since

sin 3 x + sin 2 x + sin x = sin 2 x (2 cos x + 1) = 2 sin x cos x (2 cos x + 1)

and

sin x \geq 0

[0, π]

, the substitution

t = cos x

turns the expression into

6 \int_{0}^{π} ∣ \dots ∣ d x = 12 \int_{- 1}^{1} ∣2 t^{2} + t ∣ d t

. As

2 t^{2} + t = t (2 t + 1)

is negative only on

(- \frac{1}{2}, 0)

, splitting the integral and using

\int (2 t^{2} + t) d t = \frac{2 t ^{3}}{3} + \frac{t ^{2}}{2}

gives

12 (\frac{5}{24} + \frac{1}{24} + \frac{28}{24}) = 12 \cdot \frac{34}{24} = 17

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