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 6
Q53jee_main_2025_29_jan_eveningArea Enclosed by Two Curves
Let the area enclosed between the curves |y| = 1 - x²$|y| = 1 - x^2$ and x² + y² = 1$x^2 + y^2 = 1$ be α$\alpha$. If 9α = β π + γ$9\alpha = \beta \pi + \gamma$, β, γ$\beta, \gamma$ are integers, then the value of |β - γ|$|\beta - \gamma|$ equals
A.27$27$
B.18$18$
C.15$15$
D.33$33$
Solution
Related Formula
Area enclosed between two symmetric functions across four quadrants:
α = 4 × ( Area of circle in 1st quadrant - ∫₀¹ yparabola dx )$$\alpha = 4 \times \left( \text{Area of circle in } 1^{\text{st}} \text{ quadrant} - \int_{0}^{1} y_{\text{parabola}} \, dx \right)$$
Core Logic
The curves are symmetric across the axes. The region is enclosed between the unit circle x² + y² = 1$x^2 + y^2 = 1$ and the parabolas y = ±(1 - x²)$y = \pm(1 - x^2)$.
Area Enclosed by Two Curves diagram for Q53 - JEE Main 2025 Evening
By symmetry, the area α$\alpha$ in the first quadrant is the area under the circle minus the area under the parabola from x=0$x=0$ to x=1$x=1$.
Step 1: Calculate the Area Components
Area of a circle quadrant with r=1$r=1$ is (π)/(4)$\frac{\pi}{4}$.
Symmetry helps divide the integration work by 4. Recognizing standard shapes (like a circle sector) allows you to bypass complex integration entirely for half of the problem.
Chapter Mix
Class 12 Mathematics: Integral Calculus
Q71jee_main_2025_29_jan_eveningIntegration of Modulus and Greatest Integer Functions
If 24∫₀(pi)/(4)( |4x - (pi)/(12)| + [2 x])dx = 2π +α$24\int_{0}^{\frac{pi}{4}}\left(\sin \left|4x - \frac{pi}{12}\right| + [2\sin x]\right)\mathrm{d}x = 2\pi +\alpha$, where [· ]$[\cdot ]$ denotes the greatest integer function, then α$\alpha$ is equal to
Numerical Answer.Answer: 12 to 12
Solution
Related Formula
Additivity property of definite integrals over split interval boundary points:
Splitting powers of x$x$ to create a direct match with internal derivative differential flags speeds up the integration transformation sequence.
Chapter Mix
Class 12 Mathematics: Integrals
Q67jee_main_2025_03_april_morningDefinite Integral of Greatest Integer Function
Let the domain of the function f(x) = ₂ ₄ ₆(3 + 4x - x²)$f(x) = \log_{2}\log_{4}\log_{6}(3 + 4x - x^{2})$ be (a, b)$(a, b)$. If ∫₀b-a[x²]dx = p - √(q) - √(r)$\int_{0}^{b-a}[x^{2}]dx = p - \sqrt{q} - \sqrt{r}$, where p, q, r in N$p, q, r \in \mathbb{N}$ and (p, q, r) = 1$\gcd(p, q, r) = 1$, and [·]$[\cdot]$ represents the greatest integer function, then p + q + r$p + q + r$ is equal to:
A. 10
B. 8
C. 11
D. 9
Solution
Related Formula
Domain of log chain iterations: For ₂ ₄(M) > 0$\log_2\log_4(M) > 0$, we require ₄(M) > 1 M > 4$\log_4(M) > 1 \implies M > 4$.
Compare values with requested answer template shape:
5 - √(2) - √(3) = p - √(q) - √(r)$$5 - \sqrt{2} - \sqrt{3} = p - \sqrt{q} - \sqrt{r}$$p = 5, q = 2, r = 3$$p = 5, \quad q = 2, \quad r = 3$$Final Sum = p + q + r = 5 + 2 + 3 = 10$$\text{Final Sum} = p + q + r = 5 + 2 + 3 = 10$$
Pattern Recognition
Integrals over greatest integer configurations change value exactly where the inner expression tracks through integer milestones. Mapping boundaries accurately resolves calculations smoothly.
Chapter Mix
Class 12 Mathematics: 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.