Integral Calculus – Definite Integrals – JEE Main 23 Jan 2026 Shift 2

Solution:

Let $I = \int_{0}^{x} t^2 \sin(x-t) dt$.
Use property $\int_{0}^{a} f(t) dt = \int_{0}^{a} f(a-t) dt$.
$$ I = \int_{0}^{x} (x-t)^2 \sin(t) dt $$
$$ I = \int_{0}^{x} (x^2 – 2xt + t^2) \sin t dt $$
$$ I = x^2 \int_{0}^{x} \sin t dt – 2x \int_{0}^{x} t \sin t dt + \int_{0}^{x} t^2 \sin t dt $$

Evaluating the integrals:
1) $\int_{0}^{x} \sin t dt = 1 – \cos x$
2) $\int_{0}^{x} t \sin t dt = [-t \cos t]_{0}^{x} + \int_{0}^{x} \cos t dt = -x \cos x + \sin x$
3) $\int_{0}^{x} t^2 \sin t dt = [-t^2 \cos t]_{0}^{x} + 2\int_{0}^{x} t \cos t dt = -x^2 \cos x + 2[t \sin t]_{0}^{x} – 2\int_{0}^{x} \sin t dt$
$= -x^2 \cos x + 2x \sin x – 2(1 – \cos x)$

Substitute back into expression for I:
$$ I = x^2(1 – \cos x) – 2x(-x \cos x + \sin x) + (-x^2 \cos x + 2x \sin x + 2\cos x – 2) $$
$$ I = x^2 – x^2 \cos x + 2x^2 \cos x – 2x \sin x – x^2 \cos x + 2x \sin x + 2\cos x – 2 $$
$$ I = x^2 + 2\cos x – 2 $$

Given equation: $I = x^2$.
$$ x^2 + 2\cos x – 2 = x^2 $$
$$ 2\cos x = 2 \Rightarrow \cos x = 1 $$

General solution: $x = 2n\pi, n \in Z$.
Given $x \in [0, 100]$.
$$ 0 \le 2n\pi \le 100 \Rightarrow 0 \le n \le \frac{50}{\pi} \approx 15.9 $$
Possible values for n: $0, 1, 2, …, 15$.
Total number of solutions = 16.

Ans. (16)