If ∫-(π)/(2)(π)/(2)(96x² ²x)/((1 + e^x)) dx = π (α π² +β),α ,β in Z,$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{96x^2\cos^2x}{(1 + e^x)} dx = \pi (\alpha \pi^2 +\beta),\alpha ,\beta \in \mathbb{Z},$ then (α + β)²$(\alpha + \beta)^2$ equals:

Related Formula

Core Logic

Expand the integral by distributing f(y)$f(y)$:

Since the integrals evaluate to constants, let:

Thus, the function becomes:

Step 1: Setting up equation for A

Substitute f(y) = e^y + A + B y e^y$f(y) = e^y + A + B y e^y$ back into the integral for A$A$:

Integrate by parts: ∫₀¹ y e^y dy = [y e^y - e^y]₀¹ = (e - e) - (0 - 1) = 1$\int_0^1 y e^y dy = [y e^y - e^y]_0^1 = (e - e) - (0 - 1) = 1$ ∫₀¹ y² e^y dy = [y² e^y - 2y e^y + 2e^y]₀¹ = (e - 2e + 2e) - 2 = e - 2$\int_0^1 y^2 e^y dy = [y^2 e^y - 2y e^y + 2e^y]_0^1 = (e - 2e + 2e) - 2 = e - 2$ ∫₀¹ A y dy = (A)/(2)$\int_0^1 A y dy = \frac{A}{2}$

So,

Step 2: Setting up equation for B

Substitute f(y)$f(y)$ back into the integral for B$B$:

Integrate terms: ∫₀¹ e^y dy = e - 1$\int_0^1 e^y dy = e - 1$ ∫₀¹ A dy = A$\int_0^1 A dy = A$ ∫₀¹ B y e^y dy = B(1) = B$\int_0^1 B y e^y dy = B(1) = B$

So,

0 = e - 1 + A A = 1 - e$0 = e - 1 + A \implies A = 1 - e$

Step 3: Calculating f(0)

Substitute$

Step 3: Calculating f(0)

Substitute $A = 1 - einto the expression for$ into the expression for $f(0): From the general equation,$: From the general equation, $f(x) = e^x + A + B x e^x.

Core Logic

Split the integral I$I$ at integer points and trigonometric step thresholds between -π/2 ≈ -1.57$-\pi/2 \approx -1.57$ and π/2 ≈ 1.57$\pi/2 \approx 1.57$: Intervals are: (-π/2, -1), (-1, 0), (0, 1), (1, π/2)$(-\pi/2, -1), (-1, 0), (0, 1), (1, \pi/2)$. For all x in (-π/2, π/2)$x \in (-\pi/2, \pi/2)$ excluding exactly 0$0$, [ x] = 0$[\cos x] = 0$. [ x] = -1$[\sin x] = -1$ for x in (-π/2, 0)$x \in (-\pi/2, 0)$ and [ x] = 0$[\sin x] = 0$ for x in (0, π/2)$x \in (0, \pi/2)$.

Execution

Evaluate in pieces:

• x in (-π/2, -1)$x \in (-\pi/2, -1)$:

[x] = -2, [ x] = -1, [ x] = 0$[x] = -2, [\sin x] = -1, [\cos x] = 0$. Integrand = (12(3 - 2))/(3 - 1 + 0) = (12(1))/(2) = 6$\frac{12(3 - 2)}{3 - 1 + 0} = \frac{12(1)}{2} = 6$.

• x in (-1, 0)$x \in (-1, 0)$:

[x] = -1, [ x] = -1, [ x] = 0$[x] = -1, [\sin x] = -1, [\cos x] = 0$. Integrand = (12(3 - 1))/(2) = (24)/(2) = 12$\frac{12(3 - 1)}{2} = \frac{24}{2} = 12$.

• x in (0, 1)$x \in (0, 1)$:

[x] = 0, [ x] = 0, [ x] = 0$[x] = 0, [\sin x] = 0, [\cos x] = 0$. Integrand = (12(3 + 0))/(3) = 12$\frac{12(3 + 0)}{3} = 12$.

• x in (1, π/2)$x \in (1, \pi/2)$:

[x] = 1, [ x] = 0, [ x] = 0$[x] = 1, [\sin x] = 0, [\cos x] = 0$. Integrand = (12(3 + 1))/(3) = (48)/(3) = 16$\frac{12(3 + 1)}{3} = \frac{48}{3} = 16$.

Integrate piece by piece:

I = 11π + 2$I = 11\pi + 2$

Pattern Recognition

For integrals combining integer functions and trigonometric boundaries, breaking the domain purely at integer values (-1, 0, 1)$(-1, 0, 1)$ generally perfectly matches trigonometric boundaries since the ranges of $\sin$ and $\cos$ in (-π/2, π/2)$(-\pi/2, \pi/2)$ hover near these integers.

Chapter Mix

Class 12 Maths: Definite Integration

Related Formula

King's Property of Definite Integrals:

Core Logic

Let the given integral be:

Applying King's property (x → π - x$x \to \pi - x$):

Step 1: Eliminate the x Variable

Adding equations (1) and (2):

Using the symmetric property ∫₀2a f(x)dx = 2∫₀^a f(x)dx$\int_0^{2a} f(x)dx = 2\int_0^a f(x)dx$ if f(2a-x)=f(x)$f(2a-x)=f(x)$:

Step 2: Solve Using Substitution

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:

• When x = 0 t = √(3)$x = 0 \implies t = \sqrt{3}$
• When x = π/2 t = 0$x = \pi/2 \implies t = 0$

Substituting into the integral:

Pattern Recognition

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

Related Formula

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 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$x = 1 \implies t = 2$.

Step 2: Invoke Integral Mirror Properties

Apply the identity using parameters (a+b-t) = (1-t)$(a+b-t) = (1-t)$:

Step 3: Final Matrix Matching Evaluation

Notice component blocks align exactly with I₂$I_2$ definition values:

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.

Chapter Mix

Class 12 Mathematics: Definite Integrals

Related Formula

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 (π x) ≥ 0$x \sin(\pi x) \ge 0$. For x in [1, 3/2]$x \in [1, 3/2]$, values drop below zero:

Step 2: Integrate the Even Function Block

Since x (π x)$x \sin(\pi x)$ is symmetric and even:

Step 3: Subtract the Inverse Segment

Evaluating boundary limits across the secondary phase track:

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.

Chapter Mix

Class 12 Mathematics: Definite Integrals