The value of int_-pi/6^pi/6left(fracpi+4x^111-sin(|x|+pi/6)right)dx$\int_{-\pi/6}^{\pi/6}\left(\frac{\pi+4x^{11}}{1-\sin(|x|+\pi/6)}\right)dx$ is equal to
A.2pi$2\pi$
B.4pi$4\pi$
C.8pi$8\pi$
D.6pi$6\pi$
Solution & Explanation
### Related Formula
int_-a^a f(x) dx = int_0^a [f(x) + f(-x)] dx$$\int_{-a}^{a} f(x) dx = \int_{0}^{a} [f(x) + f(-x)] dx$$
### Core Logic
Let I = int_-pi/6^pi/6fracpi+4x^111-sin(|x|+pi/6)dx$I = \int_{-\pi/6}^{\pi/6}\frac{\pi+4x^{11}}{1-\sin(|x|+\pi/6)}dx$.
The denominator 1 - sin(|x| + pi/6)$1 - \sin(|x| + \pi/6)$ is an even function.
The numerator can be split into an even part (pi$\pi$) and an odd part (4x^11$4x^{11}$).
int_-a^a frac4x^111-sin(|x|+pi/6) dx = 0 quad text(Since integrand is odd)$$\int_{-a}^{a} \frac{4x^{11}}{1-\sin(|x|+\pi/6)} dx = 0 \quad \text{(Since integrand is odd)}$$
### Step 1: Simplify to Even Integral
We are left with the even part:
I = int_-pi/6^pi/6 fracpi1 - sin(|x| + pi/6) dx$$I = \int_{-\pi/6}^{\pi/6} \frac{\pi}{1 - \sin(|x| + \pi/6)} dx$$
Using even function property int_-a^a f(x) dx = 2 int_0^a f(x) dx$\int_{-a}^{a} f(x) dx = 2 \int_{0}^{a} f(x) dx$:
I = 2pi int_0^pi/6 frac11 - sin(x + pi/6) dx$$I = 2\pi \int_{0}^{\pi/6} \frac{1}{1 - \sin(x + \pi/6)} dx$$
### Step 2: Substitution
Let t = x + fracpi6 Rightarrow dt = dx$t = x + \frac{\pi}{6} \Rightarrow dt = dx$.
Limits: when x = 0 Rightarrow t = pi/6$x = 0 \Rightarrow t = \pi/6$, when x = pi/6 Rightarrow t = pi/3$x = \pi/6 \Rightarrow t = \pi/3$.
I = 2pi int_pi/6^pi/3 fracdt1 - sin t$$I = 2\pi \int_{\pi/6}^{\pi/3} \frac{dt}{1 - \sin t}$$
### Step 3: Solve the Integral
Rationalize the denominator:
I = 2pi int_pi/6^pi/3 frac1 + sin t(1 - sin t)(1 + sin t) dt$$I = 2\pi \int_{\pi/6}^{\pi/3} \frac{1 + \sin t}{(1 - \sin t)(1 + \sin t)} dt$$I = 2pi int_pi/6^pi/3 frac1 + sin tcos^2 t dt$$I = 2\pi \int_{\pi/6}^{\pi/3} \frac{1 + \sin t}{\cos^2 t} dt$$I = 2pi int_pi/6^pi/3 (sec^2 t + sec t tan t) dt$$I = 2\pi \int_{\pi/6}^{\pi/3} (\sec^2 t + \sec t \tan t) dt$$
Integrate directly:
I = 2pi left[ tan t + sec t right]_pi/6^pi/3$$I = 2\pi \left[ \tan t + \sec t \right]_{\pi/6}^{\pi/3}$$
Evaluate limits:
Upper limit (pi/3$\pi/3$): tan(pi/3) + sec(pi/3) = sqrt3 + 2$\tan(\pi/3) + \sec(\pi/3) = \sqrt{3} + 2$
Lower limit (pi/6$\pi/6$): tan(pi/6) + sec(pi/6) = frac1sqrt3 + frac2sqrt3 = frac3sqrt3 = sqrt3$\tan(\pi/6) + \sec(\pi/6) = \frac{1}{\sqrt{3}} + \frac{2}{\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3}$I = 2pi [(sqrt3 + 2) - sqrt3] = 2pi (2) = 4pi$$I = 2\pi [(\sqrt{3} + 2) - \sqrt{3}] = 2\pi (2) = 4\pi$$
### Pattern Recognition
Symmetric limits [-a, a]$[-a, a]$ instantly demand testing for odd/even parity. Any mixed polynomial like c + k x^textodd$c + k x^{\text{odd}}$ over an even denominator guarantees the odd power term strictly vanishes, halving calculation time.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Maths: Definite Integration
Class 11 Maths: Trigonometric Functions
Keywords:#value of the integral#JEE Main 2026 Morning Q12#Definite Integration JEE Main 2026#Properties of Definite Integrals with Modulus JEE Main 2026
More Definite Integration Previous-Year Questions
Q25jee_main_2026_21_jan_morningAbsolute Value Integrals
### Related Formula
King's Property of Definite Integrals:
int_a^b f(x) dx = int_a^b f(a + b - x) dx$$\int_a^b f(x) dx = \int_a^b f(a + b - x) dx$$
### Core Logic
Let the given integral be:
I = int_0^pi frac(x + 3)sin x1 + 3cos^2x dx quad dots (1)$$I = \int_0^\pi \frac{(x + 3)\sin x}{1 + 3\cos^2x} dx \quad \dots (1)$$
Applying King's property (x to pi - x$x \to \pi - x$):
I = int_0^pi frac(pi - x + 3)sin(pi - x)1 + 3cos^2(pi - x) dx$$I = \int_0^\pi \frac{(\pi - x + 3)\sin(\pi - x)}{1 + 3\cos^2(\pi - x)} dx$$I = int_0^pi frac(pi - x + 3)sin x1 + 3cos^2x dx quad dots (2)$$I = \int_0^\pi \frac{(\pi - x + 3)\sin x}{1 + 3\cos^2x} dx \quad \dots (2)$$
### Step 1: Eliminate the x Variable
Adding equations (1) and (2):
2I = int_0^pi frac[(x + 3) + (pi - x + 3)]sin x1 + 3cos^2x dx$$2I = \int_0^\pi \frac{[(x + 3) + (\pi - x + 3)]\sin x}{1 + 3\cos^2x} dx$$2I = (pi + 6)int_0^pi fracsin x1 + 3cos^2x dx$$2I = (\pi + 6)\int_0^\pi \frac{\sin x}{1 + 3\cos^2x} dx$$
Using the symmetric property int_0^2a f(x)dx = 2int_0^a f(x)dx$\int_0^{2a} f(x)dx = 2\int_0^a f(x)dx$ if f(2a-x)=f(x)$f(2a-x)=f(x)$:
2I = 2(pi + 6)int_0^pi/2 fracsin x1 + 3cos^2x dx$$2I = 2(\pi + 6)\int_0^{\pi/2} \frac{\sin x}{1 + 3\cos^2x} dx$$I = (pi + 6)int_0^pi/2 fracsin x1 + 3cos^2x dx$$I = (\pi + 6)\int_0^{\pi/2} \frac{\sin x}{1 + 3\cos^2x} dx$$
### Step 2: Solve Using Substitution
Let t = sqrt3cos x$t = \sqrt{3}\cos x$. Then dt = -sqrt3sin x dx implies sin x dx = -fracdtsqrt3$dt = -\sqrt{3}\sin x dx \implies \sin x dx = -\frac{dt}{\sqrt{3}}$.
Change in integration boundaries:
- When x = 0 implies t = sqrt3$x = 0 \implies t = \sqrt{3}$
- When x = pi/2 implies t = 0$x = \pi/2 \implies t = 0$
Substituting into the integral:
I = (pi + 6) int_sqrt3^0 frac-dt/sqrt31 + t^2 = fracpi + 6sqrt3 int_0^sqrt3 fracdt1 + t^2$$I = (\pi + 6) \int_{\sqrt{3}}^0 \frac{-dt/\sqrt{3}}{1 + t^2} = \frac{\pi + 6}{\sqrt{3}} \int_0^{\sqrt{3}} \frac{dt}{1 + t^2}$$I = fracpi + 6sqrt3 left[ tan^-1t right]_0^sqrt3 = fracpi + 6sqrt3 left( tan^-1sqrt3 - 0 right)$$I = \frac{\pi + 6}{\sqrt{3}} \left[ \tan^{-1}t \right]_0^{\sqrt{3}} = \frac{\pi + 6}{\sqrt{3}} \left( \tan^{-1}\sqrt{3} - 0 \right)$$I = fracpi + 6sqrt3 cdot fracpi3 = fracpi3sqrt3(pi + 6)$$I = \frac{\pi + 6}{\sqrt{3}} \cdot \frac{\pi}{3} = \frac{\pi}{3\sqrt{3}}(\pi + 6)$$
### Pattern Recognition
Whenever you encounter a linear x$x$ factor multiplying trigonometric components in a definite integral with symmetric limits like 0$0$ to pi$\pi$, executing King's property first is almost guaranteed to cleanly wipe out that variable element.
### Evaluation Rubric / Model Answer
null
### 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 I_1 = int_-frac12^1 2xf(2x(1 - 2x)) \, dx$I_1 = \int_{-\frac{1}{2}}^{1} 2xf(2x(1 - 2x)) \, dx$ and I_2 = int_-1^2 f(x(1 - x)) \, dx$I_2 = \int_{-1}^{2} f(x(1 - x)) \, dx$. Then the value of fracI_2I_1$\frac{I_2}{I_1}$ is equal to
A.9$9$
B.6$6$
C.12$12$
D.4$4$
Solution
### Related Formula
int_a^b f(x) \, dx = int_a^b f(a+b-x) \, dx$$\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx$$
### Core Logic
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 I_1$I_1$, let 2x = t implies 2dx = dt$2x = t \implies 2dx = dt$.
Limits mapping: x = -1/2 implies t = -1$x = -1/2 \implies t = -1$; x = 1 implies t = 2$x = 1 \implies t = 2$.
I_1 = frac12 int_-1^2 t f(t(1-t)) \, dt implies 2I_1 = int_-1^2 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)$(a+b-t) = (1-t)$:
2I_1 = int_-1^2 (1-t) f((1-t)(1-(1-t))) \, dt$$2I_1 = \int_{-1}^{2} (1-t) f((1-t)(1-(1-t))) \, dt$$2I_1 = int_-1^2 f(t(1-t)) \, dt - int_-1^2 t f(t(1-t)) \, dt$$2I_1 = \int_{-1}^{2} f(t(1-t)) \, dt - \int_{-1}^{2} t f(t(1-t)) \, dt$$
### Step 3: Final Matrix Matching Evaluation
Notice component blocks align exactly with I_2$I_2$ definition values:
2I_1 = I_2 - 2I_1 implies 4I_1 = I_2$$2I_1 = I_2 - 2I_1 \implies 4I_1 = I_2$$fracI_2I_1 = 4$$\frac{I_2}{I_1} = 4$$
### Pattern Recognition
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.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Definite Integrals
Q61jee_main_2025_08_april_eveningIntegration of Absolute Value Functions
The integral int_-1^frac32left(left|pi^2mathrmxsin (pi mathrmx)right|right) dx$\int_{-1}^{\frac{3}{2}}\left(\left|\pi^{2}\mathrm{x}\sin (\pi \mathrm{x})\right|\right) dx$ is equal to :
A.3 + 2pi$3 + 2\pi$
B.4 + pi$4 + \pi$
C.1 + 3pi$1 + 3\pi$
D.2 + 3pi$2 + 3\pi$
Solution
### Related Formula
int x sin(pi x) \, dx = -fracxpicos(pi x) + fracsin(pi x)pi^2$$\int x \sin(\pi x) \, dx = -\frac{x}{\pi}\cos(\pi x) + \frac{\sin(\pi x)}{\pi^2}$$
### Core Logic
Track sign configurations across the target integration segments to drop absolute modulus walls effectively at clean quadrant intervals.
### Step 1: Break Apart the Modulus Domain
For x in [-1, 1]$x \in [-1, 1]$, product value elements x sin(pi x) ge 0$x \sin(\pi x) \ge 0$. For x in [1, 3/2]$x \in [1, 3/2]$, values drop below zero:
I = pi^2 left\ int_-1^1 x sin(pi x) \, dx - int_1^3/2 x sin(pi x) \, dx right\$$I = \pi^2 \left\{ \int_{-1}^{1} x \sin(\pi x) \, dx - \int_{1}^{3/2} x \sin(\pi x) \, dx \right\}$$
### Step 2: Integrate the Even Function Block
Since x sin(pi x)$x \sin(\pi x)$ is symmetric and even:
int_-1^1 x sin(pi x) \, dx = 2 int_0^1 x sin(pi x) \, dx = 2 left[ -fracxpicos(pi x) + fracsin(pi x)pi^2 right]_0^1 = frac2pi$$\int_{-1}^{1} x \sin(\pi x) \, dx = 2 \int_{0}^{1} x \sin(\pi x) \, dx = 2 \left[ -\frac{x}{\pi}\cos(\pi x) + \frac{\sin(\pi x)}{\pi^2} \right]_{0}^{1} = \frac{2}{\pi}$$
### Step 3: Subtract the Inverse Segment
Evaluating boundary limits across the secondary phase track:
int_1^3/2 x sin(pi x) \, dx = left[ -fracxpicos(pi x) + fracsin(pi x)pi^2 right]_1^3/2 = left( 0 - frac1pi^2 right) - left( frac1pi right) = -frac1pi^2 - frac1pi$$\int_{1}^{3/2} x \sin(\pi x) \, dx = \left[ -\frac{x}{\pi}\cos(\pi x) + \frac{\sin(\pi x)}{\pi^2} \right]_{1}^{3/2} = \left( 0 - \frac{1}{\pi^2} \right) - \left( \frac{1}{\pi} \right) = -\frac{1}{\pi^2} - \frac{1}{\pi}$$I = pi^2 left\ frac2pi - left(-frac1pi^2 - frac1piright) right\ = pi^2 left( frac3pi + frac1pi^2 right) = 3pi + 1$$I = \pi^2 \left\{ \frac{2}{\pi} - \left(-\frac{1}{\pi^2} - \frac{1}{\pi}\right) \right\} = \pi^2 \left( \frac{3}{\pi} + \frac{1}{\pi^2} \right) = 3\pi + 1$$
### Pattern Recognition
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.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Definite Integrals
More Definite Integration Questions — jee_main_2026_21_jan_morning
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