JEE Main 2018 (16 April Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Indefinite Integration: If $\int \frac{t a n x}{1 + t a n x + t a n ^{2} x} d x = x - \frac{K}{A} tan^{- 1} (\frac{K t a n x + 1}{A}) + C$ , (C is a constant of integration), then the ordered pair (K, A) is equal to :

Q: Wait, why do we get an $x$ in the final answer? Where did it come from?

That $x$ comes from the first part of our split integral. When we wrote the integrand as $1 - \frac{\sec^2 x}{1 + \tan x + \tan^2 x}$, we split the integral into $\int 1 dx - \int \frac{\sec^2 x}{1 + \tan x + \tan^2 x} dx$. The integral of $1 dx$ is simply $x$. It is a common point of confusion, but always remember to integrate every part of your split expression!

Select Answer:

Visualized Solution

Analyze the Integral Structure

Given integral: $I = \int \frac{t a n x}{1 + t a n x + t a n ^{2} x} d x$
Target form: $x - \frac{K}{A} tan^{- 1} (\frac{K t a n x + 1}{A}) + C$
Objective: Find the ordered pair $(K, A)$ by evaluating the integral.

Strategic Algebraic Manipulation

Manipulate the numerator to create terms similar to the denominator:
$\frac{t a n x}{1 + t a n x + t a n ^{2} x} = \frac{( 1 + t a n x + t a n ^{2} x ) - ( 1 + t a n ^{2} x )}{1 + t a n x + t a n ^{2} x}$
Recall the identity: $1 + tan^{2} x = sec^{2} x$
So, the expression becomes: $1 - \frac{s e c ^{2} x}{1 + t a n x + t a n ^{2} x}$

Splitting the Integral

Split the integral into two parts:
$I = \int 1 d x - \int \frac{s e c ^{2} x}{1 + t a n x + t a n ^{2} x} d x$
$I = x - I_{1}$ , where $I_{1} = \int \frac{s e c ^{2} x}{t a n ^{2} x + t a n x + 1} d x$

Applying Substitution

Let $u = tan x$ . Then, $d u = sec^{2} x d x$ .
Substitute these into $I_{1}$ :
$I_{1} = \int \frac{d u}{u ^{2} + u + 1}$

Completing the Square

Complete the square for $u^{2} + u + 1$ :
$u^{2} + u + 1 = (u^{2} + u + \frac{1}{4}) + 1 - \frac{1}{4}$
$u^{2} + u + 1 = (u + \frac{1}{2})^{2} + \frac{3}{4}$
Write in the form $(u + \frac{1}{2})^{2} + (\frac{3}{2})^{2}$

Standard Integration Formula

Use the formula: $\int \frac{d x}{x ^{2} + a ^{2}} = \frac{1}{a} tan^{- 1} (\frac{x}{a}) + C$
Here, $x = u + \frac{1}{2}$ and $a = \frac{3}{2}$
$I_{1} = \frac{1}{\frac{3}{2}} tan^{- 1} (\frac{u + \frac{1}{2}}{\frac{3}{2}})$
$I_{1} = \frac{2}{3} tan^{- 1} (\frac{2 u + 1}{3})$

Final Substitution and Comparison

Substitute $u = tan x$ back into the expression:
$I = x - \frac{2}{3} tan^{- 1} (\frac{2 t a n x + 1}{3}) + C$
Compare with: $x - \frac{K}{A} tan^{- 1} (\frac{K t a n x + 1}{A}) + C$
By comparison: $K = 2$ and $A = 3$
The ordered pair $(K, A)$ is $(2, 3)$ .

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The Sigma Insight: Evaluation of Special Integral Forms

Analyzing the Setup

The integral we are addressing is:

I = \int \frac{tan x}{1 + tan x + tan ^{2} x} d x

At first glance, this expression appears complex. However, in JEE Advanced mathematics, we prioritize identifying hidden structures and symmetries over brute-force calculation.

The Algebraic Insight

We begin by observing the denominator,

1 + tan x + tan^{2} x

. We aim to manipulate the numerator,

tan x

, to reflect the structure of this denominator.

By adding and subtracting terms, we can rewrite the numerator as:

tan x = (1 + tan x + tan^{2} x) - (1 + tan^{2} x)

Dividing this by the denominator yields:

\frac{( 1 + tan x + tan ^{2} x ) - ( 1 + tan ^{2} x )}{1 + tan x + tan ^{2} x} = 1 - \frac{1 + tan ^{2} x}{1 + tan x + tan ^{2} x}

Recalling the fundamental identity

1 + tan^{2} x = sec^{2} x

, the integrand simplifies to:

1 - \frac{sec ^{2} x}{1 + tan x + tan ^{2} x}

The Power of Substitution

We now split the integral into two distinct parts:

I = \int 1 d x - \int \frac{sec ^{2} x}{1 + tan x + tan ^{2} x} d x

The first part is trivial, yielding

\int 1 d x = x

. For the second part, let

I_{1} = \int \frac{s e c ^{2} x}{1 + t a n x + t a n ^{2} x} d x

Since the derivative of

tan x

sec^{2} x

, we perform the substitution

u = tan x

, which implies

d u = sec^{2} x d x

. The integral

I_{1}

transforms into:

I_{1} = \int \frac{d u}{u ^{2} + u + 1}

Completing the Square

To solve

\int \frac{d u}{u ^{2} + u + 1}

, we complete the square in the denominator:

u^{2} + u + 1 = (u^{2} + u + \frac{1}{4}) + 1 - \frac{1}{4} = (u + \frac{1}{2})^{2} + (\frac{3}{2})^{2}

Using the standard integral formula

\int \frac{d x}{x ^{2} + a ^{2}} = \frac{1}{a} tan^{- 1} (\frac{x}{a}) + C

, we obtain:

I_{1} = \frac{1}{\frac{3}{2}} tan^{- 1} (\frac{u + \frac{1}{2}}{\frac{3}{2}}) = \frac{2}{3} tan^{- 1} (\frac{2 u + 1}{3})

Final Calculation

Substituting

u = tan x

back into our expression, the complete integral is:

I = x - \frac{2}{3} tan^{- 1} (\frac{2 tan x + 1}{3}) + C

Comparing this to the target form

x - \frac{K}{A} tan^{- 1} (\frac{K t a n x + 1}{A}) + C

, we identify the constants as

K = 2

and

A = 3

The final ordered pair is $(2, 3)$ .

Common Traps & FAQs

Wait, why do we split the numerator like that? Is there a rule?

Great question! There isn't a strict rule, but it is a powerful heuristic in integration. We look for a way to make the numerator resemble the derivative of the denominator or to simplify the fraction into parts we can easily integrate. By writing

tan x

(1 + tan x + tan^{2} x) - (1 + tan^{2} x)

, we create a term that cancels out to

1

and another term that perfectly matches the derivative of

tan x

(which is

sec^{2} x

). It is all about creating symmetry.

Can't we just use partial fractions for

\frac{1}{u ^{2} + u + 1}

You could try, but you would quickly find that the roots of

u^{2} + u + 1 = 0

are complex (specifically,

ω

and

ω^{2}

). While partial fractions with complex numbers are possible, it is much more efficient to complete the square and use the standard integral formula

\int \frac{d x}{x ^{2} + a ^{2}} = \frac{1}{a} tan^{- 1} (\frac{x}{a}) + C

. It saves time and avoids unnecessary complexity.

Wait, why do we get an

x

in the final answer? Where did it come from?

That

x

comes from the first part of our split integral. When we wrote the integrand as

1 - \frac{s e c ^{2} x}{1 + t a n x + t a n ^{2} x}

, we split the integral into

\int 1 d x - \int \frac{s e c ^{2} x}{1 + t a n x + t a n ^{2} x} d x

. The integral of

1 d x

is simply

x

. It is a common point of confusion, but always remember to integrate every part of your split expression!

Animated Solution for Mathematics - Indefinite Integration: If $\int \frac{t a n x}{1 + t a n x + t a n ^{2} x} d x = x - \frac{K}{A} tan^{- 1} (\frac{K t a n x + 1}{A}) + C$ , (C is a constant of integration), then the ordered pair (K, A) is equal to :

Select Answer:

Visualized Solution

Analyze the Integral Structure

Strategic Algebraic Manipulation

Splitting the Integral

Applying Substitution

Completing the Square

Standard Integration Formula

Final Substitution and Comparison

The Sigma Insight: Evaluation of Special Integral Forms

Analyzing the Setup

The Algebraic Insight

The Power of Substitution

Completing the Square

Final Calculation

Similar Questions

If $\int \frac{2 x + 5}{7 - 6 x - x ^{2}} d x = A 7 - 6 x - x^{2} + B sin^{- 1} (\frac{x + 3}{4}) + C$ (Where C is a constant of integration), then the ordered pair (A, B) is equal to :-

If $\int \frac{5 t a n x}{t a n x - 2} d x = x + a ln ∣ sin x - 2 cos x ∣ + k$ , then $a$ is equal to

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Select Answer:

Visualized Solution

Analyze the Integral Structure

Strategic Algebraic Manipulation

Splitting the Integral

Applying Substitution

Completing the Square

Standard Integration Formula

Final Substitution and Comparison

The Sigma Insight: Evaluation of Special Integral Forms

Analyzing the Setup

The Algebraic Insight

The Power of Substitution

Completing the Square

Final Calculation

Similar Questions

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