JEE Main 2026 Jan 21 Shift 1, Mathematics Q25: Definite Integration And Fundamental Theorem

JEE Main2026Jan 21, Shift 1Mathematics

$6 \int_{0}^{π} ∣ sin 3 x + sin 2 x + sin x ∣ d x$ is equal to ……..

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Solution

Combine $sin 3 x + sin x = 2 sin 2 x cos x$ . So the integrand $= ∣ sin 2 x ∣ \cdot ∣2 cos x + 1∣ = 2∣ sin x ∣∣ cos x ∣ \cdot ∣2 cos x + 1∣$ , and on $[0, π]$ where $sin x \geq 0$ :

$I = 12 \int_{0}^{π} sin x ∣ cos x (2 cos x + 1) ∣ d x = 12 \int_{0}^{π} sin x ∣2 cos^{2} x + cos x ∣ d x$ .

Substitute $t = cos x$ , $d t = - sin x d x$ . Limits go from $1$ to $- 1$ :

$I = 12 \int_{- 1}^{1} ∣2 t^{2} + t ∣ d t$ .

$2 t^{2} + t = t (2 t + 1)$ changes sign at $t = 0$ and $t = - 1/2$ . Split the integral over $[- 1, - 1/2]$ , $[- 1/2, 0]$ , $[0, 1]$ and compute each elementary integral; the total equals $17$ .

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JEE Main 2026 Jan 21 Shift 1, Mathematics Q25: Definite Integration And Fundamental Theorem

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