Calculus – Definite Integration – JEE Main 21 Jan 2026 Shift 1

Solution:

Let $I = 6 \int_{0}^{\pi} |(\sin 3x + \sin x) + \sin 2x| dx$.

Using $\sin 3x + \sin x = 2 \sin 2x \cos x$:
$$I = 6 \int_{0}^{\pi} |2 \sin 2x \cos x + \sin 2x| dx$$
$$I = 6 \int_{0}^{\pi} |\sin 2x (2 \cos x + 1)| dx$$
$$I = 12 \int_{0}^{\pi} \sin x |2 \cos^2 x + \cos x| dx$$

Put $\cos x = t \Rightarrow -\sin x dx = dt$. Limits change from $1$ to $-1$.
$$I = 12 \int_{-1}^{1} |2t^2 + t| dt$$
Roots of $2t^2+t$ are $0, -1/2$. The expression is negative in $[-1/2, 0]$.
$$I = 12 \left[ \int_{-1}^{-1/2} (2t^2 + t) dt – \int_{-1/2}^{0} (2t^2 + t) dt + \int_{0}^{1} (2t^2 + t) dt \right]$$

$$I = 12 \left[ \left[ \frac{2t^3}{3} + \frac{t^2}{2} \right]_{-1}^{-1/2} – \left[ \frac{2t^3}{3} + \frac{t^2}{2} \right]_{-1/2}^{0} + \left[ \frac{2t^3}{3} + \frac{t^2}{2} \right]_{0}^{1} \right]$$

$$I = 12 \left[ \left( \left(\frac{-2}{24} + \frac{1}{8}\right) – \left(\frac{-2}{3} + \frac{1}{2}\right) \right) – \left( 0 – \left(\frac{-2}{24} + \frac{1}{8}\right) \right) + \left( \left(\frac{2}{3} + \frac{1}{2}\right) – 0 \right) \right]$$

$$I = 12 \left[ \left( \frac{1}{24} – \left(-\frac{1}{6}\right) \right) – \left( -\frac{1}{24} \right) + \left( \frac{7}{6} \right) \right]$$

$$I = 12 \left[ \left( \frac{1}{24} + \frac{4}{24} \right) + \frac{1}{24} + \frac{28}{24} \right]$$

$$I = 12 \left[ \frac{5}{24} + \frac{1}{24} + \frac{28}{24} \right]$$

$$I = 12 \times \frac{34}{24} = 17$$

Ans. 17