Wait, why did we only convert the $\cos^2 \theta$ in the denominator? Why not the $\cos \theta$ in the numerator?

Great observation! We are looking for a substitution. If we let $t = \sin \theta$, then $dt = \cos \theta d\theta$. We need that $\cos \theta$ in the numerator to act as the differential $dt$. If we converted it to $\sqrt{1-\sin^2 \theta}$, we would introduce square roots and make the integral significantly harder. Always look for the derivative of your substitution variable!

JEE Main 2020 (5 September Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Indefinite Integration: If $\int \frac{c o s θ}{5 + 7 s i n θ - 2 c o s ^{2} θ} d θ = A lo g_{e} ∣ B (θ) ∣ + C$ where $C$ is a constant of integration, then $\frac{B ( θ )}{A}$ can be:

Q: The question asks for $\frac{B(\theta)}{A}$. Why is $A$ the coefficient outside the log?

The problem defines the result as $A \log_e |B(\theta)| + C$. This is a standard form. Once you integrate, you get $\frac{1}{5} \log_e |...|$. By comparing, $A$ is the coefficient $\frac{1}{5}$. The function $B(\theta)$ is the argument inside the log. The ratio is simply $\frac{B(\theta)}{A}$, which is $5 \times B(\theta)$.

Select Answer:

Visualized Solution

Transforming the Denominator

Given Integral: $I = \int \frac{c o s θ}{5 + 7 s i n θ - 2 c o s ^{2} θ} d θ$
Objective: Express the denominator entirely in terms of $sin θ$ .
Use the identity: $cos^{2} θ = 1 - sin^{2} θ$

Simplifying the Expression

Substitute into the integral: $I = \int \frac{c o s θ}{5 + 7 s i n θ - 2 ( 1 - s i n ^{2} θ )} d θ$
Expand the terms: $5 + 7 sin θ - 2 + 2 sin^{2} θ$
Simplified Denominator: $2 sin^{2} θ + 7 sin θ + 3$

Applying Substitution

Let $t = sin θ$
Differentiating both sides: $cos θ d θ = d t$
Substitute $t$ and $d t$ into the integral: $I = \int \frac{d t}{2 t ^{2} + 7 t + 3}$

Factorizing the Denominator

Quadratic: $2 t^{2} + 7 t + 3$
Splitting the middle term: $2 t^{2} + 6 t + t + 3$
Factoring: $2 t (t + 3) + 1 (t + 3) = (2 t + 1) (t + 3)$
Integral becomes: $I = \int \frac{d t}{( 2 t + 1 ) ( t + 3 )}$

Partial Fraction Decomposition

$\frac{1}{( 2 t + 1 ) ( t + 3 )} = \frac{P}{2 t + 1} + \frac{Q}{t + 3}$
Multiply by denominator: $1 = P (t + 3) + Q (2 t + 1)$

Finding P and Q

To find $Q$ , let $t = - 3$ : $1 = Q (2 (- 3) + 1) ⟹ 1 = - 5 Q ⟹ Q = - \frac{1}{5}$
To find $P$ , let $t = - \frac{1}{2}$ : $1 = P (- \frac{1}{2} + 3) ⟹ 1 = \frac{5}{2} P ⟹ P = \frac{2}{5}$

Integrating the Partial Fractions

Substitute $P$ and $Q$ back: $I = \int (\frac{2/5}{2 t + 1} - \frac{1/5}{t + 3}) d t$
Factor out $\frac{1}{5}$ : $I = \frac{1}{5} \int (\frac{2}{2 t + 1} - \frac{1}{t + 3}) d t$
Integrating term by term: $I = \frac{1}{5} [lo g_{e} ∣2 t + 1∣ - lo g_{e} ∣ t + 3∣] + C$

Back-Substitution

Using $lo g A - lo g B = lo g \frac{A}{B}$ : $I = \frac{1}{5} lo g_{e} \frac{2 t + 1}{t + 3} + C$
Substitute $t = sin θ$ back: $I = \frac{1}{5} lo g_{e} \frac{2 s i n θ + 1}{s i n θ + 3} + C$

Final Comparison and Result

Comparing with $A lo g_{e} ∣ B (θ) ∣ + C$ :
$A = \frac{1}{5}$ and $B (θ) = \frac{2 s i n θ + 1}{s i n θ + 3}$
Calculate $\frac{B ( θ )}{A} = \frac{\frac{2 s i n θ + 1}{s i n θ + 3}}{1/5} = \frac{5 ( 2 s i n θ + 1 )}{s i n θ + 3}$

00:00 / 00:00

The Sigma Insight: Evaluation of Special Integral Forms

Analyzing the Setup

We begin with the integral:

I = \int \frac{cos θ}{5 + 7 sin θ - 2 cos ^{2} θ} d θ

The denominator contains both

sin θ

and

cos^{2} θ

, which creates a discord. To harmonize the expression, we apply the trigonometric identity

cos^{2} θ = 1 - sin^{2} θ

Simplifying the Denominator

Substituting the identity into the denominator, we obtain:

5 + 7 sin θ - 2 (1 - sin^{2} θ) = 2 sin^{2} θ + 7 sin θ + 3

The numerator is

cos θ d θ

, which is exactly the derivative of

sin θ

. This structure invites a direct substitution.

The Substitution Phase

Let

t = sin θ

. Consequently,

d t = cos θ d θ

. The integral transforms into:

I = \int \frac{d t}{2 t ^{2} + 7 t + 3}

We factor the quadratic expression in the denominator as

(2 t + 1) (t + 3)

Partial Fraction Decomposition

We express the integrand using partial fractions:

\frac{1}{( 2 t + 1 ) ( t + 3 )} = \frac{P}{2 t + 1} + \frac{Q}{t + 3}

Solving for the constants

P

and

Q

, we find

P = \frac{2}{5}

and

Q = - \frac{1}{5}

Final Integration

Integrating the partial fractions yields:

\frac{1}{5} ln ∣2 t + 1∣ - \frac{1}{5} ln ∣ t + 3∣ + C

Applying logarithmic properties, we simplify this to:

\frac{1}{5} ln \frac{2 t + 1}{t + 3} + C

Substituting

t = sin θ

back into the expression, we reach the final result:

I = \frac{1}{5} ln \frac{2 sin θ + 1}{sin θ + 3} + C

Comparing this to the form

A ln ∣ B (θ) ∣ + C

, we identify $A = \frac{1}{5}$ and $B (θ) = \frac{2 s i n θ + 1}{s i n θ + 3}$ . The final ratio

\frac{B ( θ )}{A}

is given by $\frac{5 ( 2 s i n θ + 1 )}{s i n θ + 3}$ .

Common Traps & FAQs

Wait, why did we only convert the

cos^{2} θ

in the denominator? Why not the

cos θ

in the numerator?

Great observation! We are looking for a substitution. If we let

t = sin θ

, then

d t = cos θ d θ

. We need that

cos θ

in the numerator to act as the differential

d t

. If we converted it to

1 - sin^{2} θ

, we would introduce square roots and make the integral significantly harder. Always look for the derivative of your substitution variable!

How do I know if I should use partial fractions or complete the square?

Excellent question. Completing the square is usually for integrals like

\int \frac{d x}{a x ^{2} + b x + c}

where the numerator is a constant and the denominator does not factor easily. Here, the denominator factors beautifully into linear terms. Partial fractions are much faster and cleaner when the denominator has distinct linear roots. Always check for factorization first!

The question asks for

\frac{B ( θ )}{A}

. Why is

A

the coefficient outside the log?

The problem defines the result as

A lo g_{e} ∣ B (θ) ∣ + C

. This is a standard form. Once you integrate, you get

\frac{1}{5} lo g_{e} ∣...∣

. By comparing,

A

is the coefficient

\frac{1}{5}

. The function

B (θ)

is the argument inside the log. The ratio is simply

\frac{B ( θ )}{A}

, which is

5 \times B (θ)

Animated Solution for Mathematics - Indefinite Integration: If $\int \frac{c o s θ}{5 + 7 s i n θ - 2 c o s ^{2} θ} d θ = A lo g_{e} ∣ B (θ) ∣ + C$ where $C$ is a constant of integration, then $\frac{B ( θ )}{A}$ can be:

Select Answer:

Visualized Solution

Transforming the Denominator

Simplifying the Expression

Applying Substitution

Factorizing the Denominator

Partial Fraction Decomposition

Finding P and Q

Integrating the Partial Fractions

Back-Substitution

Final Comparison and Result

The Sigma Insight: Evaluation of Special Integral Forms

Analyzing the Setup

Simplifying the Denominator

The Substitution Phase

Partial Fraction Decomposition

Final Integration

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Select Answer:

Visualized Solution

Transforming the Denominator

Simplifying the Expression

Applying Substitution

Factorizing the Denominator

Partial Fraction Decomposition

Finding P and Q

Integrating the Partial Fractions

Back-Substitution

Final Comparison and Result

The Sigma Insight: Evaluation of Special Integral Forms

Analyzing the Setup

Simplifying the Denominator

The Substitution Phase

Partial Fraction Decomposition

Final Integration

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