A + B = 2 ((A+B)/(2)) ((A-B)/(2))$$\sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)$$2x = 2 x x$$\sin 2x = 2 \sin x \cos x$$2x = 2 ² x - 1$$\cos 2x = 2 \cos^2 x - 1$$
Core Logic
Let I = 6∫₀π| 3x + x + 2x| dx$I = 6\int_{0}^{\pi}|\sin 3x + \sin x + \sin 2x| dx$.
Apply sum-to-product on 3x + x$\sin 3x + \sin x$:
3x + x = 2 (2x) (x)$\sin 3x + \sin x = 2 \sin(2x) \cos(x)$
So the expression becomes:
|2 (2x) x + 2x| = | 2x (2 x + 1)| = |2 x x (2 x + 1)|$|2 \sin(2x) \cos x + \sin 2x| = |\sin 2x (2 \cos x + 1)| = |2 \sin x \cos x (2 \cos x + 1)|$
Since x in [0, π]$x \in [0, \pi]$, x ≥ 0$\sin x \geq 0$. We can pull it out of the modulus.
I = 12 ∫₀π x |2 ² x + x| dx$I = 12 \int_{0}^{\pi} \sin x |2 \cos^2 x + \cos x| dx$
Step 1: Coordinate Substitution
Substitute t = x$t = \cos x$, then dt = - x dx$dt = -\sin x dx$.
Limits: when x = 0$x = 0$, t = 1$t = 1$. When x = π$x = \pi$, t = -1$t = -1$.
The roots of 2t² + t = 0$2t^2 + t = 0$ are t = 0$t = 0$ and t = -1/2$t = -1/2$.
The quadratic 2t² + t$2t^2 + t$ is negative in the interval (-1/2, 0)$(-1/2, 0)$ and positive elsewhere.
Split the integral:
Whenever an integral features a cascading sum of sine frequencies like (kx)$\sin(kx)$, pair the highest and lowest frequencies first. The resulting common factor often matches the middle term, instantly yielding a clean polynomial substitution under t = x$t = \cos x$.
Chapter Mix
Class 12 Maths: Definite Integration
Class 11 Maths: Trigonometric Functions
Keywords:#is equal to#JEE Main 2026 Morning Q25#Definite Integration JEE Main 2026#Absolute Value Integrals JEE Main 2026
More Definite Integration Previous-Year Questions — Page 5
Q71jee_main_2025_02_april_morningDefinite Integration with GIF
Let [·]$[\cdot]$ denote the greatest integer function. If ∫₀^e³[ 1ex - 1]dx = α - ₑ2$\int_{0}^{e^{3}}\left[\frac{1}{e^{x - 1}}\right]\mathrm{d}x = \alpha -\log_{e}2$, then α³$\alpha^3$ is equal to ________.
Numerical Answer.Answer: 8 to 8
Solution
Related Formula
Greatest Integer Function boundaries:
[f(x)] = k for k ≤ f(x) < k+1, k in Z$$[f(x)] = k \quad \text{for} \quad k \le f(x) < k+1, \ k \in \mathbb{Z}$$
Core Logic
Analyze the value variations of f(x) = e1-x$f(x) = e^{1-x}$ across the integration limits [0, e³]$[0, e^3]$ to break down the integral into distinct piecewise continuous intervals.
Step 1: Determine Step Function Transition Points
Let y = e1-x$y = e^{1-x}$.
At x = 0 y = e¹ ≈ 2.718$x = 0 \implies y = e^1 \approx 2.718$
As x$x$ increases, e1-x$e^{1-x}$ decreases monotonically.
Find x$x$ where y = 2 e1-x = 2 1-x = ln 2 x = 1 - ln 2$y = 2 \implies e^{1-x} = 2 \implies 1-x = \ln 2 \implies x = 1 - \ln 2$.
Find x$x$ where y = 1 e1-x = 1 1-x = 0 x = 1$y = 1 \implies e^{1-x} = 1 \implies 1-x = 0 \implies x = 1$.
At the final boundary x = e³ y = e1-e³$x = e^3 \implies y = e^{1-e^3}$, which is a very small positive decimal strictly inside (0,1)$(0,1)$.
Step 2: Split the Definite Integral
Rewrite the integral based on the isolated interval blocks:
Always map the function values at the extreme boundary points first. Tracking the downward path from 2.71 arrow 2 arrow 1 arrow 0$2.71 \rightarrow 2 \rightarrow 1 \rightarrow 0$ reveals exactly where the integer thresholds are crossed.
The presence of a linear x$x$ factor in the numerator of a definite integral with symmetric trigonometric bounds is almost always eliminated using the a+b-x$a+b-x$ property. This reduces the integral to a standard substitution problem.
Chapter Mix
Class 12 Mathematics: Integrals
Q60jee_main_2025_07_april_morningProperties of Definite Integrals
The integral ∫₀^π ((x + 3) x)/(1 + 3 ²x) dx$\int_0^\pi \frac{(x + 3)\sin x}{1 + 3\cos^2x} dx$ is equal to:
Let t = √(3) x$t = \sqrt{3}\cos x$. Then dt = -√(3) x dx x dx = - dt√(3)$dt = -\sqrt{3}\sin x dx \implies \sin x dx = -\frac{dt}{\sqrt{3}}$.
Change in integration boundaries:
Whenever you encounter a linear x$x$ factor multiplying trigonometric components in a definite integral with symmetric limits like 0$0$ to π$\pi$, executing King's property first is almost guaranteed to cleanly wipe out that variable element.
Chapter Mix
Class 12 Mathematics: Definite Integrals
Q58jee_main_2025_08_april_eveningProperties of Definite Integrals
Let f(x)$f(x)$ be a a positive function and I₁ = ∫-(1)/(2)¹ 2xf(2x(1 - 2x)) dx$I_1 = \int_{-\frac{1}{2}}^{1} 2xf(2x(1 - 2x)) \, dx$ and I₂ = ∫₋₁² f(x(1 - x)) dx$I_2 = \int_{-1}^{2} f(x(1 - x)) \, dx$. Then the value of (I₂)/(I₁)$\frac{I_2}{I_1}$ is equal to
Perform variable substitution to match the arguments and limit bounds across both separate integral functions before invoking King's property.
Step 1: Perform Base Transformation Substitution
In I₁$I_1$, let 2x = t 2dx = dt$2x = t \implies 2dx = dt$.
Limits mapping: x = -1/2 t = -1$x = -1/2 \implies t = -1$; x = 1 t = 2$x = 1 \implies t = 2$.
I₁ = (1)/(2) ∫₋₁² t f(t(1-t)) dt 2I₁ = ∫₋₁² t f(t(1-t)) dt$$I_1 = \frac{1}{2} \int_{-1}^{2} t f(t(1-t)) \, dt \implies 2I_1 = \int_{-1}^{2} t f(t(1-t)) \, dt$$
Step 2: Invoke Integral Mirror Properties
Apply the identity using parameters (a+b-t) = (1-t)$(a+b-t) = (1-t)$:
Symmetric transformations highlighting factor expressions like x(1-x)$x(1-x)$ coupled with an external linear multiplier term x$x$ naturally simplify to half-weight area forms using reflection rules.
Chapter Mix
Class 12 Mathematics: Definite Integrals
Q61jee_main_2025_08_april_eveningIntegration of Absolute Value Functions
The integral ∫₋₁(3)/(2)(|π²x (π x)|) dx$\int_{-1}^{\frac{3}{2}}\left(\left|\pi^{2}\mathrm{x}\sin (\pi \mathrm{x})\right|\right) dx$ is equal to :
Products of two odd tracking metrics (like linear variable x$x$ matched with sinusoidal waves) yield overall even systems, enabling rapid evaluation over center-aligned domains.
Chapter Mix
Class 12 Mathematics: Definite Integrals
More Definite Integration Questions — jee_main_2026_21_jan_morning
Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.