Why do the limits change from $[0, \pi]$ to $[0, \infty)$?

Think about the behavior of the tangent function. At $x = 0$, $\tan(0/2) = 0$. As $x$ approaches $\pi$, the argument of the tangent function, $x/2$, approaches $\pi/2$. We know that $\tan(\theta)$ approaches infinity as $\theta$ approaches $\pi/2$. So, the entire interval $[0, \pi]$ is mapped onto the entire positive real line $[0, \infty)$. It is a beautiful transformation of space!

JEE Main 2024 (27 Jan Shift 2)

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Animated Solution for Mathematics - Definite Integration: For $0 < a < 1$ , the value of the integral $\int_{0}^{π} \frac{d x}{1 - 2 a c o s x + a ^{2}}$ is :

Q: Wait, why do we use $t = \tan(x/2)$? Can't we just use a simpler substitution like $u = \cos x$?

That is a fantastic question! If you try $u = \cos x$, you get $du = -\sin x dx$. But look at our integral: we have a $dx$ in the numerator, not a $\sin x dx$. You would be left with a $\sin x$ term in the denominator, which is $\sqrt{1-u^2}$. This leads to a messy square root integral. The Weierstrass substitution $t = \tan(x/2)$ is the 'nuclear option' because it converts the entire trigonometric expression into a rational algebraic one, which we know how to solve perfectly!

Q: What if $a > 1$? Does the answer change?

You are thinking like a true mathematician! If $a > 1$, the denominator $1 - 2a \cos x + a^2$ could potentially become zero for some values of $x$. Specifically, if $a > 1$, the function would have a singularity, and the integral would diverge. The constraint $0 < a < 1$ is not just a random condition; it is the 'safety zone' that ensures our integral is well-defined and finite.

Select Answer:

Progressive Hints

Hint 1

The denominator

1 - 2 a cos x + a^{2}

is a classic structure. Before diving into calculus, look at the constraint

0 < a < 1

. Does this guarantee the denominator is never zero? Visualize the function; it is smooth and continuous, inviting us to use a powerful transformation.

Hint 2

When you see a trigonometric function of

x

in the denominator, especially one that looks like a quadratic in

cos x

, the Weierstrass substitution is your best friend. Recall the substitution

t = tan (\frac{x}{2})

. This is the 'universal key' that unlocks trigonometric integrals by converting them into rational algebraic functions.

Hint 3

Start by setting

t = tan (\frac{x}{2})

. Remember that

d x = \frac{2 d t}{1 + t ^{2}}

and

cos x = \frac{1 - t ^{2}}{1 + t ^{2}}

. Substitute these into your integral and watch the limits transform:

x = 0

becomes

t = 0

, and

x = π

becomes

t \to \infty

. Your integral will transform into

\int_{0}^{\infty} \frac{2 d t}{( 1 + t ^{2} ) - 2 a ( 1 - t ^{2} ) + a ^{2} ( 1 + t ^{2} )}

Visualized Solution

Visualizing the Integral

Given Integral: $I = \int_{0}^{π} \frac{d x}{1 - 2 a c o s x + a ^{2}}$
Constraint: $0 < a < 1$
Objective: Evaluate the definite integral over the interval $[0, π]$

Choosing the Substitution

Let $t = tan (\frac{x}{2})$
This is the standard Weierstrass Substitution for trigonometric integrals.

Finding $d x$ in terms of $d t$

Differentiating $t = tan (\frac{x}{2})$ :
$d t = \frac{1}{2} sec^{2} (\frac{x}{2}) d x$
$d t = \frac{1}{2} (1 + tan^{2} (\frac{x}{2})) d x$
$d t = \frac{1}{2} (1 + t^{2}) d x$
$⟹ d x = \frac{2 d t}{1 + t ^{2}}$

Expressing $cos x$ in terms of $t$

Using the half-angle identity:
$cos x = \frac{1 - t a n ^{2} ( x /2 )}{1 + t a n ^{2} ( x /2 )}$
$⟹ cos x = \frac{1 - t ^{2}}{1 + t ^{2}}$

Changing the Limits

Lower limit: When $x = 0$ , $t = tan (0) = 0$
Upper limit: When $x \to π$ , $t = tan (\frac{π}{2}) \to \infty$
New limits for $t$ : $[0, \infty)$

The Raw Substitution

Substitute $d x$ and $cos x$ into $I$ :
$I = \int_{0}^{\infty} \frac{\frac{2 d t}{1 + t ^{2}}}{1 - 2 a ( \frac{1 - t ^{2}}{1 + t ^{2}} ) + a ^{2}}$

Simplifying the Denominator

Multiply numerator and denominator by $(1 + t^{2})$ :
$I = \int_{0}^{\infty} \frac{2 d t}{( 1 + t ^{2} ) - 2 a ( 1 - t ^{2} ) + a ^{2} ( 1 + t ^{2} )}$

Expanding and Grouping Terms

Expand the terms in the denominator:
$1 + t^{2} - 2 a + 2 a t^{2} + a^{2} + a^{2} t^{2}$
Group terms of $t^{2}$ and constants:
$I = \int_{0}^{\infty} \frac{2 d t}{( 1 + 2 a + a ^{2} ) t ^{2} + ( 1 - 2 a + a ^{2} )}$

Recognizing Perfect Squares

Notice the perfect squares:
$1 + 2 a + a^{2} = (1 + a)^{2}$
$1 - 2 a + a^{2} = (1 - a)^{2}$
Substitute back:
$I = \int_{0}^{\infty} \frac{2 d t}{( 1 + a ) ^{2} t ^{2} + ( 1 - a ) ^{2}}$

Preparing for the Standard Formula

Factor out $(1 + a)^{2}$ from the denominator to isolate $t^{2}$ :
$I = \frac{2}{( 1 + a ) ^{2}} \int_{0}^{\infty} \frac{d t}{t ^{2} + ( \frac{1 - a}{1 + a} ) ^{2}}$

Applying the Standard Formula

Standard Integral: $\int \frac{d u}{u ^{2} + k ^{2}} = \frac{1}{k} tan^{- 1} (\frac{u}{k})$
Here $k = \frac{1 - a}{1 + a}$
Applying the formula:
$I = \frac{2}{( 1 + a ) ^{2}} \cdot \frac{1}{\frac{1 - a}{1 + a}} [tan^{- 1} (\frac{t}{\frac{1 - a}{1 + a}})]_{0}^{\infty}$

Simplifying the Coefficients

Simplify the constant factor outside the bracket:
$\frac{2}{( 1 + a ) ^{2}} \cdot \frac{1 + a}{1 - a} = \frac{2}{( 1 + a ) ( 1 - a )} = \frac{2}{1 - a ^{2}}$
So, $I = \frac{2}{1 - a ^{2}} [tan^{- 1} (\frac{1 + a}{1 - a} t)]_{0}^{\infty}$

Evaluating the Limits

Upper limit: As $t \to \infty$ , $tan^{- 1} (\infty) = \frac{π}{2}$
Lower limit: As $t = 0$ , $tan^{- 1} (0) = 0$
Note: Since $0 < a < 1$ , the factor $\frac{1 + a}{1 - a}$ is positive.
Calculation:
$I = \frac{2}{1 - a ^{2}} (\frac{π}{2} - 0)$

The Final Result

Multiply the terms:
$I = \frac{2}{1 - a ^{2}} \cdot \frac{π}{2}$
Final Answer: $I = \frac{π}{1 - a ^{2}}$
This matches the given option.

00:00 / 00:00

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Solution Diagram

Analyzing the Setup

We are tasked with evaluating the integral:

I = \int_{0}^{π} \frac{d x}{1 - 2 a cos x + a ^{2}}

The constraint

0 < a < 1

is our guiding star, ensuring that our denominator remains positive and our function remains smooth throughout the interval.

The Universal Key

Weierstrass Substitution

When you encounter a rational function of

sin x

and

cos x

, the Weierstrass substitution is the most powerful tool available. We set

t = tan (\frac{x}{2})

This substitution acts as a bridge between the circular world of trigonometry and the linear world of algebra. By setting

t = tan (\frac{x}{2})

, we utilize the following identities:

d x = \frac{2 d t}{1 + t ^{2}} and cos x = \frac{1 - t ^{2}}{1 + t ^{2}}

The Transformation of Limits

Now, we must transform the boundaries of our integral. Our original integral spans from

x = 0

x = π

x = 0

t = tan (0) = 0

. As

x \to π

t \to tan (π /2)

, which approaches

\infty

. Our integral now stretches across the entire positive real line:

I = \int_{0}^{\infty} \frac{\frac{2 d t}{1 + t ^{2}}}{1 - 2 a ( \frac{1 - t ^{2}}{1 + t ^{2}} ) + a ^{2}}

The Algebraic Revelation

To simplify, we multiply the numerator and the denominator by

(1 + t^{2})

. The numerator becomes

2 d t

, and the denominator transforms into:

(1 + t^{2}) - 2 a (1 - t^{2}) + a^{2} (1 + t^{2})

Grouping the terms by

t^{2}

and constants, we obtain

(1 + 2 a + a^{2}) t^{2} + (1 - 2 a + a^{2})

. Recognizing these as perfect squares, we have

(1 + a)^{2} t^{2} + (1 - a)^{2}

. The integral becomes:

I = \int_{0}^{\infty} \frac{2 d t}{( 1 + a ) ^{2} t ^{2} + ( 1 - a ) ^{2}}

The Final Victory

We factor out

(1 + a)^{2}

to normalize the coefficient of

t^{2}

I = \frac{2}{( 1 + a ) ^{2}} \int_{0}^{\infty} \frac{d t}{t ^{2} + ( \frac{1 - a}{1 + a} ) ^{2}}

Using the standard integral form

\int \frac{d u}{u ^{2} + k ^{2}} = \frac{1}{k} tan^{- 1} (\frac{u}{k})

, we evaluate:

I = \frac{2}{( 1 + a ) ^{2}} \cdot \frac{1 + a}{1 - a} [tan^{- 1} (\frac{1 + a}{1 - a} t)]_{0}^{\infty}

t \to \infty

, the

tan^{- 1}

term approaches

π /2

, and at

t = 0

, it is

0

. Simplifying the constants, we find:

I = \frac{2}{( 1 + a ) ( 1 - a )} \cdot \frac{π}{2} = \frac{π}{1 - a ^{2}}

The final result is:

I = \frac{π}{1 - a ^{2}}

Common Traps & FAQs

Wait, why do we use

t = tan (x /2)

? Can't we just use a simpler substitution like

u = cos x

That is a fantastic question! If you try

u = cos x

, you get

d u = - sin x d x

. But look at our integral: we have a

d x

in the numerator, not a

sin x d x

. You would be left with a

sin x

term in the denominator, which is

1 - u^{2}

. This leads to a messy square root integral. The Weierstrass substitution

t = tan (x /2)

is the 'nuclear option' because it converts the entire trigonometric expression into a rational algebraic one, which we know how to solve perfectly!

Why do the limits change from

[0, π]

[0, \infty)

Think about the behavior of the tangent function. At

x = 0

tan (0/2) = 0

. As

x

approaches

π

, the argument of the tangent function,

x /2

, approaches

π /2

. We know that

tan (θ)

approaches infinity as

θ

approaches

π /2

. So, the entire interval

[0, π]

is mapped onto the entire positive real line

[0, \infty)

. It is a beautiful transformation of space!

What if

a > 1

? Does the answer change?

You are thinking like a true mathematician! If

a > 1

, the denominator

1 - 2 a cos x + a^{2}

could potentially become zero for some values of

x

. Specifically, if

a > 1

, the function would have a singularity, and the integral would diverge. The constraint

0 < a < 1

is not just a random condition; it is the 'safety zone' that ensures our integral is well-defined and finite.

Animated Solution for Mathematics - Definite Integration: For $0 < a < 1$ , the value of the integral $\int_{0}^{π} \frac{d x}{1 - 2 a c o s x + a ^{2}}$ is :

Select Answer:

Visualized Solution

Visualizing the Integral

Choosing the Substitution

Finding $d x$ in terms of $d t$

Expressing $cos x$ in terms of $t$

Changing the Limits

The Raw Substitution

Simplifying the Denominator

Expanding and Grouping Terms

Recognizing Perfect Squares

Preparing for the Standard Formula

Applying the Standard Formula

Simplifying the Coefficients

Evaluating the Limits

The Final Result

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

The Universal Key

The Transformation of Limits

The Algebraic Revelation

The Final Victory

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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; } For 0<a<1, the value of the integral ∫0π​1−2acosx+a2dx​ is :

Select Answer:

Visualized Solution

Visualizing the Integral

Choosing the Substitution

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } dx in terms of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } dt

Expressing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } cosx in terms of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t

Changing the Limits

The Raw Substitution

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Expanding and Grouping Terms

Recognizing Perfect Squares

Preparing for the Standard Formula

Applying the Standard Formula

Simplifying the Coefficients

Evaluating the Limits

The Final Result

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

The Universal Key

The Transformation of Limits

The Algebraic Revelation

The Final Victory

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Animated Solution for Mathematics - Definite Integration: For $0 < a < 1$ , the value of the integral $\int_{0}^{π} \frac{d x}{1 - 2 a c o s x + a ^{2}}$ is :

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