JEE Main 2020 (9 January Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Indefinite Integration: If $\int \frac{d θ}{c o s ^{2} θ ( t a n 2 θ + s e c 2 θ )} = λ tan θ + 2 lo g_{e} ∣ f (θ) ∣ + C$ where $C$ is a constant of integration, then the ordered pair $(λ, f (θ))$ is equal to:

Select Answer:

Visualized Solution

The Integral Structure

Given integral: $I = \int \frac{d θ}{c o s ^{2} θ ( t a n 2 θ + s e c 2 θ )}$
Target form: $λ tan θ + 2 lo g_{e} ∣ f (θ) ∣ + C$

Double Angle Formulas

Recall: $tan 2 θ = \frac{2 t a n θ}{1 - t a n ^{2} θ}$
Recall: $sec 2 θ = \frac{1}{c o s 2 θ} = \frac{1 + t a n ^{2} θ}{1 - t a n ^{2} θ}$

Denominator Substitution

Substitute into denominator: $tan 2 θ + sec 2 θ = \frac{2 t a n θ}{1 - t a n ^{2} θ} + \frac{1 + t a n ^{2} θ}{1 - t a n ^{2} θ}$

Simplifying the Denominator

Combine fractions: $\frac{1 + 2 t a n θ + t a n ^{2} θ}{1 - t a n ^{2} θ}$
Identify perfect square: $\frac{( 1 + t a n θ ) ^{2}}{1 - t a n ^{2} θ}$

Integral in terms of $tan θ$

Rewrite integral: $I = \int \frac{s e c ^{2} θ d θ}{\frac{( 1 + t a n θ ) ^{2}}{1 - t a n ^{2} θ}}$
Simplify: $I = \int \frac{s e c ^{2} θ ( 1 - t a n ^{2} θ )}{( 1 + t a n θ ) ^{2}} d θ$

Substitution $t = tan θ$

Let $t = tan θ$
Then $d t = sec^{2} θ d θ$

Algebraic Form

Substitute into integral: $I = \int \frac{1 - t ^{2}}{( 1 + t ) ^{2}} d t$

Factoring the Numerator

Factorize: $1 - t^{2} = (1 - t) (1 + t)$
Cancel common term: $I = \int \frac{( 1 - t ) ( 1 + t )}{( 1 + t ) ^{2}} d t = \int \frac{1 - t}{1 + t} d t$

Splitting the Fraction

Rewrite numerator: $1 - t = 2 - (1 + t)$
Split fraction: $\frac{1 - t}{1 + t} = \frac{2}{1 + t} - \frac{1 + t}{1 + t} = \frac{2}{1 + t} - 1$

Integration

Integrate: $I = \int (\frac{2}{1 + t} - 1) d t$
Result: $I = 2 lo g_{e} ∣1 + t ∣ - t + C$

Back Substitution

Substitute $t = tan θ$ : $I = 2 lo g_{e} ∣1 + tan θ ∣ - tan θ + C$
Rearrange: $I = - 1 \cdot tan θ + 2 lo g_{e} ∣1 + tan θ ∣ + C$

Final Comparison

Compare with $λ tan θ + 2 lo g_{e} ∣ f (θ) ∣ + C$
$λ = - 1$
$f (θ) = 1 + tan θ$
Final Answer: $(- 1, 1 + tan θ)$

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The Sigma Insight: Integration by Substitution

Analyzing the Setup

The integral we are tasked to solve is:

I = \int \frac{d θ}{cos ^{2} θ ( tan 2 θ + sec 2 θ )}

In JEE Advanced, intimidation is often a sign that you need a better perspective. Our goal is to reach the target form

λ tan θ + 2 lo g_{e} ∣ f (θ) ∣ + C

The Bridge

Double Angle Identities

The integrand contains

tan 2 θ

and

sec 2 θ

. To simplify these, we apply the standard double angle identities:

tan 2 θ = \frac{2 tan θ}{1 - tan ^{2} θ}, sec 2 θ = \frac{1 + tan ^{2} θ}{1 - tan ^{2} θ}

Notice that both expressions share the same denominator,

1 - tan^{2} θ

. Adding them together yields:

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

The Magic of Substitution

Recalling that

\frac{1}{c o s ^{2} θ} = sec^{2} θ

, we substitute our simplified expression back into the integral:

I = \int \frac{sec ^{2} θ ( 1 - tan ^{2} θ )}{( 1 + tan θ ) ^{2}} d θ

We now perform the substitution

t = tan θ

, which implies

d t = sec^{2} θ d θ

. The integral transforms into a purely algebraic form:

I = \int \frac{1 - t ^{2}}{( 1 + t ) ^{2}} d t

The Algebraic Cleanup

The numerator

1 - t^{2}

is a difference of squares,

(1 - t) (1 + t)

. Canceling the common

(1 + t)

term, we obtain:

I = \int \frac{1 - t}{1 + t} d t

To integrate this, we rewrite the numerator as

2 - (1 + t)

I = \int (\frac{2}{1 + t} - 1) d t = 2 lo g_{e} ∣1 + t ∣ - t + C

The Final Victory

Substituting

t = tan θ

back into the expression, we get:

I = 2 lo g_{e} ∣1 + tan θ ∣ - tan θ + C

Comparing this to the target form

λ tan θ + 2 lo g_{e} ∣ f (θ) ∣ + C

, we identify the parameters:

$λ = - 1$ and $f (θ) = 1 + tan θ$ .

The final result is the ordered pair $(- 1, 1 + tan θ)$ .

Common Traps & FAQs

Wait, why do we convert everything to

tan θ

? Can't we just use

sin θ

and

cos θ

You absolutely could, but it would be a path of immense suffering! The target form

λ tan θ + 2 lo g_{e} ∣ f (θ) ∣ + C

is your compass. It explicitly tells you that the final answer is a function of

tan θ

. By converting early, you align your strategy with the destination, making the integration path much smoother.

Why did we split the fraction

\frac{1 - t}{1 + t}

into

\frac{2}{1 + t} - 1

That is a classic algebraic trick! When you have a rational function where the degree of the numerator is equal to the degree of the denominator, you should always perform long division or a simple adjustment to separate the constant part from the fractional part. This makes the integration trivial because

\int \frac{1}{1 + t} d t

is a standard logarithmic form.

Is it possible to solve this without the substitution

t = tan θ

While you could technically manipulate the trigonometric functions directly, the substitution

t = tan θ

is the 'golden key' for integrals involving

sec^{2} θ

and

tan θ

. It transforms a complex trigonometric integral into a simple rational algebraic one, which is much less prone to error.

Animated Solution for Mathematics - Indefinite Integration: If $\int \frac{d θ}{c o s ^{2} θ ( t a n 2 θ + s e c 2 θ )} = λ tan θ + 2 lo g_{e} ∣ f (θ) ∣ + C$ where $C$ is a constant of integration, then the ordered pair $(λ, f (θ))$ is equal to:

Select Answer:

Visualized Solution

The Integral Structure

Double Angle Formulas

Denominator Substitution

Simplifying the Denominator

Integral in terms of $tan θ$

Substitution $t = tan θ$

Algebraic Form

Factoring the Numerator

Splitting the Fraction

Integration

Back Substitution

Final Comparison

The Sigma Insight: Integration by Substitution

Analyzing the Setup

The Bridge

The Magic of Substitution

The Algebraic Cleanup

The Final Victory

Similar Questions

If $\int \frac{d θ}{c o s ^{2} θ ( t a n 2 θ + s e c 2 θ )} = λ tan θ + 2 lo g_{e} ∣ f (θ) ∣ + C$ where C is constant if integration, then the ordered pair $(λ, f (θ))$ is equal to:

If $\int sin^{- 1} (\frac{x}{1 + x}) d x = A (x) tan^{- 1} (x) + B (x) + C$ where $C$ is a constant of integration, then the ordered pair $(A (x), B (x))$ can be

If $\int \frac{c o s x - s i n x}{8 - s i n 2 x} d x = a sin^{- 1} (\frac{s i n x + c o s x}{b}) + c$ , where $c$ is a constant of integration, then the ordered pair $(a, b)$ is equal to :

Let $α \in (0, π /2)$ be fixed. If the integral $\int \frac{t a n x + t a n α}{t a n x - t a n α} d x = A (x) cos 2 α + B (x) sin 2 α + C$ , where C is a constant of integration, then the functions A(x) and B(x) are respectively :

If $\int \frac{c o s x d x}{s i n ^{2} x ( 1 + s i n x ) ^{2/3}} = f (x) (1 + sin^{6} x)^{1/ λ} + c$ where $c$ is a constant of integration, then $λ f (\frac{π}{3})$ is equal to :

If $f (\frac{x - 4}{x + 2}) = 2 x + 1, (x \in R - {1, - 2})$ , then $\int f (x) d x$ is equal to : (where C is a constant of integration)

For $x^{2} \neq = nπ + 1$ , $n \in N$ (the set of natural numbers), the integral $\int \frac{2 s i n ( x ^{2} - 1 ) - s i n 2 ( x ^{2} - 1 )}{2 s i n ( x ^{2} - 1 ) + s i n 2 ( x ^{2} - 1 )} d x$ is equal to : (where c is a constant of integration)

If $\int sec 2 x - 1 d x = α lo g_{e} cos 2 x + β + cos 2 x (1 + cos \frac{1}{β} x) + constant$ , then $β - α$ is equal to ______.

If $\int \frac{1}{x} \frac{1 - x}{1 + x} d x = g (x) + c, g (1) = 0$ , then $g (\frac{1}{2})$ is equal to :

The integral $\int \frac{s e c ^{2} x}{( s e c x + t a n x ) ^{9/2}} d x$ equals (for some arbitrary constant $K$ )

Select Answer:

Visualized Solution

The Integral Structure

Double Angle Formulas

Denominator Substitution

Simplifying the Denominator

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

Substitution .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t=tanθ

Algebraic Form

Factoring the Numerator

Splitting the Fraction

Integration

Back Substitution

Final Comparison

The Sigma Insight: Integration by Substitution

Analyzing the Setup

The Bridge

The Magic of Substitution

The Algebraic Cleanup

The Final Victory

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(x+2x−4​)=2x+1,(x∈R−{1,−2}), then ∫f(x)dx is equal to : (where C is a constant of integration)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If ∫sec2x−1​dx=αloge​​cos2x+β+cos2x(1+cosβ1​x)​​+constant, then β−α is equal to ______.

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Integral in terms of $tan θ$

Substitution $t = tan θ$

If $\int \frac{d θ}{c o s ^{2} θ ( t a n 2 θ + s e c 2 θ )} = λ tan θ + 2 lo g_{e} ∣ f (θ) ∣ + C$ where C is constant if integration, then the ordered pair $(λ, f (θ))$ is equal to:

If $\int sin^{- 1} (\frac{x}{1 + x}) d x = A (x) tan^{- 1} (x) + B (x) + C$ where $C$ is a constant of integration, then the ordered pair $(A (x), B (x))$ can be

If $\int \frac{c o s x - s i n x}{8 - s i n 2 x} d x = a sin^{- 1} (\frac{s i n x + c o s x}{b}) + c$ , where $c$ is a constant of integration, then the ordered pair $(a, b)$ is equal to :

If $\int \frac{c o s x d x}{s i n ^{2} x ( 1 + s i n x ) ^{2/3}} = f (x) (1 + sin^{6} x)^{1/ λ} + c$ where $c$ is a constant of integration, then $λ f (\frac{π}{3})$ is equal to :

If $f (\frac{x - 4}{x + 2}) = 2 x + 1, (x \in R - {1, - 2})$ , then $\int f (x) d x$ is equal to : (where C is a constant of integration)

If $\int sec 2 x - 1 d x = α lo g_{e} cos 2 x + β + cos 2 x (1 + cos \frac{1}{β} x) + constant$ , then $β - α$ is equal to ______.

If $\int \frac{1}{x} \frac{1 - x}{1 + x} d x = g (x) + c, g (1) = 0$ , then $g (\frac{1}{2})$ is equal to :

The integral $\int \frac{s e c ^{2} x}{( s e c x + t a n x ) ^{9/2}} d x$ equals (for some arbitrary constant $K$ )