JEE Advanced 2016

LEVELJEE Main

Animated Solution for Mathematics - Definite Integration: The value of $\int_{- π /2}^{π /2} \frac{x ^{2} c o s x}{1 + e ^{x}} d x$ is equal to

Q: Wait, why do we use the King's Property here? Can't we just integrate by parts directly?

That is a fantastic question! If you attempt to integrate $\frac{x^2 \cos x}{1 + e^x}$ directly, you will hit a wall because the denominator $1 + e^x$ has no simple antiderivative in terms of elementary functions. The King's Property is a strategic tool; it is designed to eliminate the 'problematic' part of the integrand by creating a symmetric partner that cancels it out. Always look for the denominator that refuses to be integrated as a sign to use this property!

Q: Why does the denominator $1 + e^{-x}$ become $1 + \frac{1}{e^x}$ and then simplify to $\frac{e^x}{1 + e^x}$?

This is a classic algebraic hurdle. When you substitute $-x$ for $x$, you get $e^{-x}$. Writing this as $\frac{1}{e^x}$ is the first step. Then, to simplify the complex fraction $\frac{1}{1 + \frac{1}{e^x}}$, you multiply the numerator and the denominator by $e^x$. This gives you $\frac{e^x}{e^x(1 + \frac{1}{e^x})}$, which simplifies to $\frac{e^x}{e^x + 1}$. It is a standard algebraic trick to clear denominators!

Q: Why is $x^2 \cos x$ an even function?

A function $f(x)$ is even if $f(-x) = f(x)$. Let's check: $f(-x) = (-x)^2 \cos(-x)$. Since $(-x)^2 = x^2$ and $\cos(-x) = \cos x$ (because cosine is an even function), we get $x^2 \cos x$, which is exactly $f(x)$. Since the function is even and the limits are symmetric ($-\frac{\pi}{2}$ to $\frac{\pi}{2}$), we can simplify the integral to $2 \int_{0}^{\pi/2} f(x) dx$.

Select Answer:

Progressive Hints

Hint 1

Hint 1: Observe the limits of integration:

[- \frac{π}{2}, \frac{π}{2}]

. Whenever you encounter symmetric limits like

[- a, a]

, your mathematical intuition should immediately trigger the use of the King's Property. This property is specifically designed to dismantle complex denominators that appear impossible to integrate directly.

Hint 2

Hint 2: Recall the King's Property:

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

. In this problem,

a = - \frac{π}{2}

and

b = \frac{π}{2}

, so

a + b = 0

. This means you are essentially replacing

x

with

- x

. Consider how the term

\frac{1}{1 + e ^{x}}

transforms when you replace

x

with

- x

Hint 3

Hint 3: Once you have the transformed integral, add it to the original integral

I

. You will discover that the denominator

1 + e^{x}

magically cancels out when you combine the numerators. You will be left with a much simpler integral:

2 I = \int_{- π /2}^{π /2} x^{2} cos x d x

. Now, apply the property of even functions to solve this.

Visualized Solution

Introduction to the Integral $I$

Let the given integral be $I = \int_{- π /2}^{π /2} \frac{x ^{2} c o s x}{1 + e ^{x}} d x$
Observe the limits: $[- \frac{π}{2}, \frac{π}{2}]$ . This suggests using properties of definite integrals.

Applying King's Property

Apply the property: $\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x$
Here, $a = - \frac{π}{2}$ and $b = \frac{π}{2}$ , so $a + b = 0$ .
Replace $x$ with $(0 - x) = - x$ .

Substituting $- x$ into the Integrand

$I = \int_{- π /2}^{π /2} \frac{( - x ) ^{2} c o s ( - x )}{1 + e ^{- x}} d x$
Since $(- x)^{2} = x^{2}$ and $cos (- x) = cos x$ :
$I = \int_{- π /2}^{π /2} \frac{x ^{2} c o s x}{1 + e ^{- x}} d x$

Simplifying the New Expression

Rewrite $e^{- x}$ as $\frac{1}{e ^{x}}$ :
$I = \int_{- π /2}^{π /2} \frac{x ^{2} c o s x}{1 + \frac{1}{e ^{x}}} d x$
Taking LCM in the denominator: $I = \int_{- π /2}^{π /2} \frac{e ^{x} x ^{2} c o s x}{1 + e ^{x}} d x$

Adding the Two Integrals

$2 I = \int_{- π /2}^{π /2} \frac{x ^{2} c o s x}{1 + e ^{x}} d x + \int_{- π /2}^{π /2} \frac{e ^{x} x ^{2} c o s x}{1 + e ^{x}} d x$
$2 I = \int_{- π /2}^{π /2} \frac{x ^{2} c o s x ( 1 + e ^{x} )}{1 + e ^{x}} d x$
$2 I = \int_{- π /2}^{π /2} x^{2} cos x d x$

Using Symmetry Properties

The integrand $f (x) = x^{2} cos x$ is an even function because $f (- x) = f (x)$ .
Property: $\int_{- a}^{a} f (x) d x = 2 \int_{0}^{a} f (x) d x$ for even functions.

Simplifying to Half Interval

$2 I = 2 \int_{0}^{π /2} x^{2} cos x d x ⟹ I = \int_{0}^{π /2} x^{2} cos x d x$

Integration by Parts: Step 1

Use Integration by Parts (IBP): $\int u d v = uv - \int v d u$
Let $u = x^{2} ⟹ d u = 2 x d x$
Let $d v = cos x d x ⟹ v = sin x$

Integration by Parts: Step 2

$I = [x^{2} sin x]_{0}^{π /2} - \int_{0}^{π /2} 2 x sin x d x$

Integration by Parts: Step 3

Apply IBP again to $\int 2 x sin x d x$ :
Let $u = 2 x ⟹ d u = 2 d x$
Let $d v = sin x d x ⟹ v = - cos x$
Result: $- 2 x cos x + 2 sin x$

Combining the Results

$I = [x^{2} sin x - (- 2 x cos x + 2 sin x)]_{0}^{π /2}$
$I = [x^{2} sin x + 2 x cos x - 2 sin x]_{0}^{π /2}$

Evaluating the Upper Limit

At $x = \frac{π}{2}$ :
$(\frac{π}{2})^{2} sin (\frac{π}{2}) + 2 (\frac{π}{2}) cos (\frac{π}{2}) - 2 sin (\frac{π}{2})$
$= \frac{π ^{2}}{4} (1) + π (0) - 2 (1) = \frac{π ^{2}}{4} - 2$

Evaluating the Lower Limit

At $x = 0$ :
$(0)^{2} sin (0) + 2 (0) cos (0) - 2 sin (0) = 0 + 0 - 0 = 0$

Final Conclusion

$I = \frac{π ^{2}}{4} - 2$
Key Takeaway: King's Property $\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x$ eliminates exponential denominators for symmetric limits.

00:00 / 00:00

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Solution Diagram

Analyzing the Setup

The integral

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

presents a formidable challenge due to the exponential term in the denominator. In JEE Advanced mathematics, such structures are rarely solved via standard antiderivatives.

The symmetric limits

[- \frac{π}{2}, \frac{π}{2}]

serve as a critical signal to employ the King's Property.

The King's Property

The Secret Weapon

The King's Property states that:

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

Here,

a = - \frac{π}{2}

and

b = \frac{π}{2}

, implying

a + b = 0

. We replace every

x

with

- x

I = \int_{- π /2}^{π /2} \frac{( - x ) ^{2} cos ( - x )}{1 + e ^{- x}} d x

Since

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

and

cos (- x) = cos x

, the numerator remains unchanged. We rewrite the denominator

1 + e^{- x}

1 + \frac{1}{e ^{x}} = \frac{e ^{x} + 1}{e ^{x}}

Multiplying the numerator and denominator by

e^{x}

, the integral transforms into:

I = \int_{- π /2}^{π /2} \frac{e ^{x} x ^{2} cos x}{1 + e ^{x}} d x

The Beautiful Cancellation

We now possess two versions of the same integral. Adding them together yields:

2 I = \int_{- π /2}^{π /2} \frac{x ^{2} cos x}{1 + e ^{x}} d x + \int_{- π /2}^{π /2} \frac{e ^{x} x ^{2} cos x}{1 + e ^{x}} d x

Combining the numerators results in

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

. The

(1 + e^{x})

terms cancel out perfectly:

2 I = \int_{- π /2}^{π /2} x^{2} cos x d x

Symmetry and Integration by Parts

Since

x^{2} cos x

is an even function, we simplify the integral:

2 I = 2 \int_{0}^{π /2} x^{2} cos x d x \Rightarrow I = \int_{0}^{π /2} x^{2} cos x d x

We apply Integration by Parts with

u = x^{2}

and

d v = cos x d x

, giving

d u = 2 x d x

and

v = sin x

I = [x^{2} sin x]_{0}^{π /2} - \int_{0}^{π /2} 2 x sin x d x

Applying Integration by Parts again to

\int 2 x sin x d x

(where

u = 2 x

and

d v = sin x d x

I = [x^{2} sin x]_{0}^{π /2} - ([- 2 x cos x]_{0}^{π /2} + \int_{0}^{π /2} 2 cos x d x)

Evaluating the terms:

I = [x^{2} sin x + 2 x cos x - 2 sin x]_{0}^{π /2}

Substituting the limits

\frac{π}{2}

and

0

I = (\frac{π ^{2}}{4} (1) + 2 (\frac{π}{2}) (0) - 2 (1)) - (0 + 0 - 0)

The final result is:

I = \frac{π ^{2}}{4} - 2

Common Traps & FAQs

Wait, why do we use the King's Property here? Can't we just integrate by parts directly?

That is a fantastic question! If you attempt to integrate

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

directly, you will hit a wall because the denominator

1 + e^{x}

has no simple antiderivative in terms of elementary functions. The King's Property is a strategic tool; it is designed to eliminate the 'problematic' part of the integrand by creating a symmetric partner that cancels it out. Always look for the denominator that refuses to be integrated as a sign to use this property!

Why does the denominator

1 + e^{- x}

become

1 + \frac{1}{e ^{x}}

and then simplify to

\frac{e ^{x}}{1 + e ^{x}}

This is a classic algebraic hurdle. When you substitute

- x

for

x

, you get

e^{- x}

. Writing this as

\frac{1}{e ^{x}}

is the first step. Then, to simplify the complex fraction

\frac{1}{1 + \frac{1}{e ^{x}}}

, you multiply the numerator and the denominator by

e^{x}

. This gives you

\frac{e ^{x}}{e ^{x} ( 1 + \frac{1}{e ^{x}} )}

, which simplifies to

\frac{e ^{x}}{e ^{x} + 1}

. It is a standard algebraic trick to clear denominators!

Why is

x^{2} cos x

an even function?

A function

f (x)

is even if

f (- x) = f (x)

. Let's check:

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

. Since

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

and

cos (- x) = cos x

(because cosine is an even function), we get

x^{2} cos x

, which is exactly

f (x)

. Since the function is even and the limits are symmetric (

- \frac{π}{2}

\frac{π}{2}

), we can simplify the integral to

2 \int_{0}^{π /2} f (x) d x

Animated Solution for Mathematics - Definite Integration: The value of $\int_{- π /2}^{π /2} \frac{x ^{2} c o s x}{1 + e ^{x}} d x$ is equal to

Select Answer:

Visualized Solution

Introduction to the Integral $I$

Applying King's Property

Substituting $- x$ into the Integrand

Simplifying the New Expression

Adding the Two Integrals

Using Symmetry Properties

Simplifying to Half Interval

Integration by Parts: Step 1

Integration by Parts: Step 2

Integration by Parts: Step 3

Combining the Results

Evaluating the Upper Limit

Evaluating the Lower Limit

Final Conclusion

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

The King's Property

The Beautiful Cancellation

Symmetry and Integration by Parts

Similar Questions

The value of $\int_{0}^{π} \frac{e ^{c o s x} s i n x}{( 1 + c o s ^{2} x ) ( e ^{c o s x} + e ^{- c o s x} )} d x$ is equal to

The value of $\int_{- π /2}^{π /2} (\frac{1 + s i n ^{2} x}{1 + π ^{s i n x}}) d x$ is

The value of $\int_{- π}^{π} \frac{c o s ^{2} x}{1 + a ^{x}} d x, a > 0$ , is

The value of $\int_{- π}^{π} \frac{c o s ^{2} x}{1 + a ^{x}} d x, a > 0$ , is

The value of the definite integral $\int_{- π /4}^{π /4} \frac{d x}{( 1 + e ^{x c o s x} ) ( s i n ^{4} x + c o s ^{4} x )}$ is equal to:

The value of the integral $\int_{- π /2}^{π /2} (x^{2} + ln \frac{π + x}{π - x}) cos x d x$ is

The value of $\int_{- π /2}^{π /2} \frac{s i n ^{2} x}{1 + 2 ^{x}} d x$ is :

The value of $\int_{e^{2}}^{e^{4}} \frac{1}{x} (\frac{e ^{((l o g_{e} x)^{2} + 1)^{- 1}}}{e ^{((l o g_{e} x)^{2} + 1)^{- 1}} + e ^{((6 - l o g_{e} x)^{2} + 1)^{- 1}}}) d x$ is

Integrate $\int_{0}^{π} \frac{e ^{c o s x}}{e ^{c o s x} + e ^{- c o s x}} d x$ .

Determine the value of $\int_{- π}^{π} \frac{2 x ( 1 + s i n x )}{1 + c o s ^{2} x} d x$ .

Animated Solution for Mathematics - Definite Integration: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of ∫−π/2π/2​1+exx2cosx​dx is equal to

Select Answer:

Visualized Solution

Introduction to the Integral .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I

Applying King's Property

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } −x into the Integrand

Simplifying the New Expression

Adding the Two Integrals

Using Symmetry Properties

Simplifying to Half Interval

Integration by Parts: Step 1

Integration by Parts: Step 2

Integration by Parts: Step 3

Combining the Results

Evaluating the Upper Limit

Evaluating the Lower Limit

Final Conclusion

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

The King's Property

The Beautiful Cancellation

Symmetry and Integration by Parts

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Animated Solution for Mathematics - Definite Integration: The value of $\int_{- π /2}^{π /2} \frac{x ^{2} c o s x}{1 + e ^{x}} d x$ is equal to

Introduction to the Integral $I$

Substituting $- x$ into the Integrand

The value of $\int_{0}^{π} \frac{e ^{c o s x} s i n x}{( 1 + c o s ^{2} x ) ( e ^{c o s x} + e ^{- c o s x} )} d x$ is equal to

The value of $\int_{- π /2}^{π /2} (\frac{1 + s i n ^{2} x}{1 + π ^{s i n x}}) d x$ is

The value of $\int_{- π}^{π} \frac{c o s ^{2} x}{1 + a ^{x}} d x, a > 0$ , is

The value of $\int_{- π}^{π} \frac{c o s ^{2} x}{1 + a ^{x}} d x, a > 0$ , is

The value of the definite integral $\int_{- π /4}^{π /4} \frac{d x}{( 1 + e ^{x c o s x} ) ( s i n ^{4} x + c o s ^{4} x )}$ is equal to:

The value of the integral $\int_{- π /2}^{π /2} (x^{2} + ln \frac{π + x}{π - x}) cos x d x$ is

The value of $\int_{- π /2}^{π /2} \frac{s i n ^{2} x}{1 + 2 ^{x}} d x$ is :

The value of $\int_{e^{2}}^{e^{4}} \frac{1}{x} (\frac{e ^{((l o g_{e} x)^{2} + 1)^{- 1}}}{e ^{((l o g_{e} x)^{2} + 1)^{- 1}} + e ^{((6 - l o g_{e} x)^{2} + 1)^{- 1}}}) d x$ is

Integrate $\int_{0}^{π} \frac{e ^{c o s x}}{e ^{c o s x} + e ^{- c o s x}} d x$ .

Determine the value of $\int_{- π}^{π} \frac{2 x ( 1 + s i n x )}{1 + c o s ^{2} x} d x$ .