$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{96x^{2}\cos^{2}x}{(1+e^{x})}dx=\pi(\alpha\pi^{2}+\beta)$ , $\alpha,\beta\in\mathbb{Z}$ then $(\alpha+\beta)^{2}$ equals

### Analyzing the Setup To evaluate the integral $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{96x^2 \cos^2 x}{1+e^x} dx$, we utilize the symmetry property of definite integrals: $$\int_{-a}^{a} f(x) dx = \int_{0}^{a} (f(x) + f(-x)) dx$$ Let $f(x) = \frac{96x^2 \cos^2 x}{1+e^x}$. Observing the components, we note that $(-x)^2 = x^2$ and $\cos(-x) = \cos x$, meaning the numerator is an even function. When we calculate $f(-x)$, we obtain: $$f(-x) = \frac{96x^2 \cos^2 x}{1+e^{-x}} = \frac{96x^2 \cos^2 x \cdot e^x}{e^x + 1}$$ Summing $f(x)$ and $f(-x)$ yields: $$f(x) + f(-x) = \frac{96x^2 \cos^2 x}{1+e^x} + \frac{96x^2 \cos^2 x \cdot e^x}{1+e^x} = 96x^2 \cos^2 x \left( \frac{1+e^x}{1+e^x} \right) = 96x^2 \cos^2 x$$ ### The Trigonometric Bridge The integral simplifies to $I = \int_{0}^{\frac{\pi}{2}} 96x^2 \cos^2 x dx$. Applying the identity $\cos^2 x = \frac{1+\cos 2x}{2}$, we rewrite the expression as: $$I = 48 \int_{0}^{\frac{\pi}{2}} x^2 dx + 48 \int_{0}^{\frac{\pi}{2}} x^2 \cos 2x dx$$ The first part is straightforward: $$48 \left[ \frac{x^3}{3} \right]_{0}^{\frac{\pi}{2}} = 16 \left( \frac{\pi^3}{8} \right) = 2\pi^3$$ For the second part, $48 \int_{0}^{\frac{\pi}{2}} x^2 \cos 2x dx$, we use Integration by Parts with $u = x^2$ and $dv = \cos 2x dx$: $$48 \left( \left[ x^2 \frac{\sin 2x}{2} \right]_{0}^{\frac{\pi}{2}} - \int_{0}^{\frac{\pi}{2}} 2x \frac{\sin 2x}{2} dx \right) = 48 \left( 0 - \int_{0}^{\frac{\pi}{2}} x \sin 2x dx \right) = -48 \int_{0}^{\frac{\pi}{2}} x \sin 2x dx$$ ### The Final Reduction We apply Integration by Parts again to $\int_{0}^{\frac{\pi}{2}} x \sin 2x dx$, setting $u = x$ and $dv = \sin 2x dx$: $$-48 \left( \left[ -x \frac{\cos 2x}{2} \right]_{0}^{\frac{\pi}{2}} - \int_{0}^{\frac{\pi}{2}} -\frac{\cos 2x}{2} dx \right)$$ Evaluating the boundary terms: $$-48 \left( -\frac{\pi}{2} \cdot \frac{\cos \pi}{2} - 0 + \frac{1}{2} \left[ \frac{\sin 2x}{2} \right]_{0}^{\frac{\pi}{2}} \right) = -48 \left( \frac{\pi}{4} + 0 \right) = -12\pi$$ Combining the results, we find $I = 2\pi^3 - 12\pi = \pi(2\pi^2 - 12)$. Comparing this to $\pi(\alpha\pi^2 + \beta)$, we identify $\alpha = 2$ and $\beta = -12$. The final result is: $$(\alpha + \beta)^2 = (2 - 12)^2 = (-10)^2 = \mathbf{100}$$

JEE Main 2025 (January)

LEVELJEE Main

Animated Solution for Mathematics - Definite Integration: $\int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{96 x ^{2} c o s ^{2} x}{( 1 + e ^{x} )} d x = π (α π^{2} + β)$ , $α, β \in Z$ then $(α + β)^{2}$ equals

Q: Why do we have to perform Integration by Parts twice? Can't we just do it once?

Great question! When you have an integral of the form $\int x^n \cos(ax) dx$, you need to use IBP $n$ times to completely eliminate the $x^n$ term. In our case, we have $x^2$. The first IBP reduces $x^2$ to $x$, and the second IBP reduces $x$ to a constant. If you stop after one step, you are still left with an integral involving $x$, which you cannot solve directly. Persistence is key here!

Enter Numerical Value:

Visualized Solution

Understanding the Problem

Given Integral: $I = \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{96 x ^{2} c o s ^{2} x}{1 + e ^{x}} d x$
Target: Find $(α + β)^{2}$ where $I = π (α π^{2} + β)$

Applying the Symmetry Property

Property: $\int_{- a}^{a} f (x) d x = \int_{0}^{a} (f (x) + f (- x)) d x$
Here $f (x) = \frac{96 x ^{2} c o s ^{2} x}{1 + e ^{x}}$

Finding $f (- x)$

$f (- x) = \frac{96 ( - x ) ^{2} c o s ^{2} ( - x )}{1 + e ^{- x}}$
Since $(- x)^{2} = x^{2}$ and $cos (- x) = cos x$ :
$f (- x) = \frac{96 x ^{2} c o s ^{2} x}{1 + e ^{- x}}$

Simplifying $f (x) + f (- x)$

$f (x) + f (- x) = 96 x^{2} cos^{2} x (\frac{1}{1 + e ^{x}} + \frac{1}{1 + e ^{- x}})$
Using $\frac{1}{1 + e ^{- x}} = \frac{e ^{x}}{e ^{x} + 1}$ :
$f (x) + f (- x) = 96 x^{2} cos^{2} x$

The New Integral

New Integral: $I = \int_{0}^{\frac{π}{2}} 96 x^{2} cos^{2} x d x$
Using $cos^{2} x = \frac{1 + c o s 2 x}{2}$ :
$I = \int_{0}^{\frac{π}{2}} 96 x^{2} (\frac{1 + c o s 2 x}{2}) d x$

Splitting the Integral

$I = 48 \int_{0}^{\frac{π}{2}} (x^{2} + x^{2} cos 2 x) d x$
Splitting: $I = 48 \int_{0}^{\frac{π}{2}} x^{2} d x + 48 \int_{0}^{\frac{π}{2}} x^{2} cos 2 x d x$

Solving the First Integral

First Part: $48 [\frac{x ^{3}}{3}]_{0}^{\frac{π}{2}}$
$= 16 [x^{3}]_{0}^{\frac{π}{2}} = 16 (\frac{π}{2})^{3}$
$= 16 \cdot \frac{π ^{3}}{8} = 2 π^{3}$

Integration by Parts - Step 1

Second Part: $48 \int_{0}^{\frac{π}{2}} x^{2} cos 2 x d x$
Using IBP: $\int u d v = uv - \int v d u$
Let $u = x^{2} ⟹ d u = 2 x d x$
Let $d v = cos 2 x d x ⟹ v = \frac{s i n 2 x}{2}$

Applying First IBP

$= 48 ([x^{2} \frac{s i n 2 x}{2}]_{0}^{\frac{π}{2}} - \int_{0}^{\frac{π}{2}} 2 x \frac{s i n 2 x}{2} d x)$
Evaluate boundary term: $(\frac{π}{2})^{2} \frac{s i n π}{2} - 0 = 0$
Simplifies to: $- 48 \int_{0}^{\frac{π}{2}} x sin 2 x d x$

Integration by Parts - Step 2

Integral: $- 48 \int_{0}^{\frac{π}{2}} x sin 2 x d x$
Apply IBP again: $u = x ⟹ d u = d x$
$d v = sin 2 x d x ⟹ v = - \frac{c o s 2 x}{2}$

Evaluating the Second IBP

$= - 48 ([- x \frac{c o s 2 x}{2}]_{0}^{\frac{π}{2}} - \int_{0}^{\frac{π}{2}} (- \frac{c o s 2 x}{2}) d x)$
$= - 48 ((- \frac{π}{2} \frac{c o s π}{2} - 0) + \int_{0}^{\frac{π}{2}} \frac{c o s 2 x}{2} d x)$
$= - 48 (\frac{π}{4} + [\frac{s i n 2 x}{4}]_{0}^{\frac{π}{2}})$

Finalizing the Second Part

Evaluate $[\frac{s i n 2 x}{4}]_{0}^{\frac{π}{2}} = \frac{s i n π}{4} - \frac{s i n 0}{4} = 0$
Second Part Value: $- 48 (\frac{π}{4} + 0) = - 12 π$

Combining the Results

Total Integral $I = 2 π^{3} - 12 π$
Factoring out $π$ : $I = π (2 π^{2} - 12)$
Comparing with given form: $I = π (α π^{2} + β)$
Therefore, $α = 2$ and $β = - 12$

Final Calculation

We have $α = 2$ and $β = - 12$
Calculate $α + β = 2 - 12 = - 10$
Final Value: $(α + β)^{2} = (- 10)^{2} = 100$

00:00 / 00:00

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

To evaluate the integral

I = \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{96 x ^{2} c o s ^{2} x}{1 + e ^{x}} d x

, we utilize the symmetry property of definite integrals:

\int_{- a}^{a} f (x) d x = \int_{0}^{a} (f (x) + f (- x)) d x

Let

f (x) = \frac{96 x ^{2} c o s ^{2} x}{1 + e ^{x}}

. Observing the components, we note that

(- x)^{2} = x^{2}

and

cos (- x) = cos x

, meaning the numerator is an even function.

When we calculate

f (- x)

, we obtain:

f (- x) = \frac{96 x ^{2} cos ^{2} x}{1 + e ^{- x}} = \frac{96 x ^{2} cos ^{2} x \cdot e ^{x}}{e ^{x} + 1}

Summing

f (x)

and

f (- x)

yields:

f (x) + f (- x) = \frac{96 x ^{2} cos ^{2} x}{1 + e ^{x}} + \frac{96 x ^{2} cos ^{2} x \cdot e ^{x}}{1 + e ^{x}} = 96 x^{2} cos^{2} x (\frac{1 + e ^{x}}{1 + e ^{x}}) = 96 x^{2} cos^{2} x

The Trigonometric Bridge

The integral simplifies to

I = \int_{0}^{\frac{π}{2}} 96 x^{2} cos^{2} x d x

. Applying the identity

cos^{2} x = \frac{1 + c o s 2 x}{2}

, we rewrite the expression as:

I = 48 \int_{0}^{\frac{π}{2}} x^{2} d x + 48 \int_{0}^{\frac{π}{2}} x^{2} cos 2 x d x

The first part is straightforward:

48 [\frac{x ^{3}}{3}]_{0}^{\frac{π}{2}} = 16 (\frac{π ^{3}}{8}) = 2 π^{3}

For the second part,

48 \int_{0}^{\frac{π}{2}} x^{2} cos 2 x d x

, we use Integration by Parts with

u = x^{2}

and

d v = cos 2 x d x

48 ([x^{2} \frac{sin 2 x}{2}]_{0}^{\frac{π}{2}} - \int_{0}^{\frac{π}{2}} 2 x \frac{sin 2 x}{2} d x) = 48 (0 - \int_{0}^{\frac{π}{2}} x sin 2 x d x) = - 48 \int_{0}^{\frac{π}{2}} x sin 2 x d x

The Final Reduction

We apply Integration by Parts again to

\int_{0}^{\frac{π}{2}} x sin 2 x d x

, setting

u = x

and

d v = sin 2 x d x

- 48 ([- x \frac{cos 2 x}{2}]_{0}^{\frac{π}{2}} - \int_{0}^{\frac{π}{2}} - \frac{cos 2 x}{2} d x)

Evaluating the boundary terms:

- 48 (- \frac{π}{2} \cdot \frac{cos π}{2} - 0 + \frac{1}{2} [\frac{sin 2 x}{2}]_{0}^{\frac{π}{2}}) = - 48 (\frac{π}{4} + 0) = - 12 π

Combining the results, we find

I = 2 π^{3} - 12 π = π (2 π^{2} - 12)

. Comparing this to

π (α π^{2} + β)

, we identify

α = 2

and

β = - 12

The final result is:

(α + β)^{2} = (2 - 12)^{2} = (- 10)^{2} = 100

Common Traps & FAQs

Wait, why did the denominator

1 + e^{x}

just disappear? That feels like magic.

It is not magic, but a beautiful property of definite integrals! When you evaluate

f (x) + f (- x)

, you get

\frac{96 x ^{2} c o s ^{2} x}{1 + e ^{x}} + \frac{96 x ^{2} c o s ^{2} x}{1 + e ^{- x}}

. If you multiply the numerator and denominator of the second term by

e^{x}

, it becomes

\frac{96 x ^{2} c o s ^{2} x \cdot e ^{x}}{e ^{x} + 1}

. Now, both terms have the same denominator

(1 + e^{x})

, and the numerators add up to

96 x^{2} cos^{2} x (1 + e^{x})

. The

(1 + e^{x})

terms cancel out perfectly. It is a classic JEE trap designed to test if you recognize this symmetry!

Why do we have to perform Integration by Parts twice? Can't we just do it once?

Great question! When you have an integral of the form

\int x^{n} cos (a x) d x

, you need to use IBP

n

times to completely eliminate the

x^{n}

term. In our case, we have

x^{2}

. The first IBP reduces

x^{2}

x

, and the second IBP reduces

x

to a constant. If you stop after one step, you are still left with an integral involving

x

, which you cannot solve directly. Persistence is key here!

Is there a faster way to solve

\int x^{2} cos 2 x d x

without doing IBP twice?

Yes, there is a technique called the Tabular Method (or DI Method). You create two columns: one for differentiation (D) and one for integration (I). You differentiate

x^{2}

until you reach zero (

x^{2} \to 2 x \to 2 \to 0

) and integrate

cos 2 x

the same number of times. Then, you multiply diagonally with alternating signs. It is a fantastic shortcut for JEE Advanced, but make sure you practice it enough to avoid sign errors!

Animated Solution for Mathematics - Definite Integration: $\int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{96 x ^{2} c o s ^{2} x}{( 1 + e ^{x} )} d x = π (α π^{2} + β)$ , $α, β \in Z$ then $(α + β)^{2}$ equals

Enter Numerical Value:

Visualized Solution

Understanding the Problem

Applying the Symmetry Property

Finding $f (- x)$

Simplifying $f (x) + f (- x)$

The New Integral

Splitting the Integral

Solving the First Integral

Integration by Parts - Step 1

Applying First IBP

Integration by Parts - Step 2

Evaluating the Second IBP

Finalizing the Second Part

Combining the Results

Final Calculation

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

The Trigonometric Bridge

The Final Reduction

Similar Questions

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Animated Solution for Mathematics - Definite Integration: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∫−2π​2π​​(1+ex)96x2cos2x​dx=π(απ2+β) , α,β∈Z then (α+β)2 equals

Enter Numerical Value:

Visualized Solution

Understanding the Problem

Applying the Symmetry Property

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(−x)

Simplifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)+f(−x)

The New Integral

Splitting the Integral

Solving the First Integral

Integration by Parts - Step 1

Applying First IBP

Integration by Parts - Step 2

Evaluating the Second IBP

Finalizing the Second Part

Combining the Results

Final Calculation

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

The Trigonometric Bridge

The Final Reduction

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Animated Solution for Mathematics - Definite Integration: $\int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{96 x ^{2} c o s ^{2} x}{( 1 + e ^{x} )} d x = π (α π^{2} + β)$ , $α, β \in Z$ then $(α + β)^{2}$ equals

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