Why is the final answer $8 \csc 2\theta$ and not just $4 \csc 2\theta$?

Don't forget the constants! When you differentiate $2 \sec \theta \sqrt{\dots}$, the chain rule brings out a factor of $1/\cos \theta$. When you combine the two parts, you get $2 \sec \theta$ and $2 \csc \theta$. Multiplying these gives $4 \sec \theta \csc \theta$, which is $4 / (\sin \theta \cos \theta)$. Multiplying by $2/2$ gives $8 / (2 \sin \theta \cos \theta)$, which is $8 \csc 2\theta$.

JEE Main 2024 (29 Jan Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Indefinite Integration: If $\int \frac{s i n ^{3/2} x + c o s ^{3/2} x}{s i n ^{3} x c o s ^{3} x s i n ( x - θ )} d x = A cos θ tan x - sin θ + B cos θ - sin θ cot x + C$ where C is the integration constant, then AB is equal to

Q: Why do we use $t^2$ instead of just $t$ for the substitution?

Using $t^2$ is a classic trick for integrals involving square roots. By setting the entire radical equal to $t$, you get $t = \sqrt{\dots}$, which is fine, but setting the *inside* of the radical to $t^2$ allows the square root to vanish immediately, leaving you with a simple linear integral. It saves you from dealing with messy fractional powers during the integration process!

Select Answer:

Progressive Hints

Hint 1

Hint 1: The Spark. Look at the denominator. The term

sin (x - θ)

is the key to unlocking the entire expression. If you expand it using the compound angle identity

sin x cos θ - cos x sin θ

, you will see that you can factor out either

cos x

sin x

from the radical to create perfect differentials for

tan x

cot x

Hint 2

Hint 2: The Arsenal. Once you have split the integral into two parts,

I_{1}

and

I_{2}

, focus on the structure of the differentials. For

I_{1}

, you need to create a

sec^{2} x

term to facilitate a substitution with

tan x

. For

I_{2}

, you need a

csc^{2} x

term to facilitate a substitution with

cot x

Hint 3

Hint 3: The Execution. Start by setting

I_{1} = \int \frac{s e c ^{2} x}{t a n x c o s θ - s i n θ} d x

and

I_{2} = \int \frac{c s c ^{2} x}{c o s θ - c o t x s i n θ} d x

. Use the substitution

u^{2} = tan x cos θ - sin θ

for the first and

v^{2} = cos θ - cot x sin θ

for the second.

Visualized Solution

Analyzing the Integral Structure

Given integral: $I = \int \frac{s i n ^{\frac{3}{2}} x + c o s ^{\frac{3}{2}} x}{s i n ^{3} x c o s ^{3} x s i n ( x - θ )} d x$
Target form: $A cos θ tan x - sin θ + B cos θ - sin θ cot x + C$
Strategy: Expand the compound angle and split the integral into two parts to match the target terms.

Expanding $sin (x - θ)$

Using the identity: $sin (x - θ) = sin x cos θ - cos x sin θ$
Substitute into the integral:
$I = \int \frac{s i n ^{\frac{3}{2}} x + c o s ^{\frac{3}{2}} x}{s i n ^{3} x c o s ^{3} x \cdot ( s i n x c o s θ - c o s x s i n θ )} d x$

Splitting the Integral into $I_{1}$ and $I_{2}$

Split the integral: $I = I_{1} + I_{2}$
$I_{1} = \int \frac{s i n ^{\frac{3}{2}} x}{s i n ^{3} x c o s ^{3} x ( s i n x c o s θ - c o s x s i n θ )} d x$
$I_{2} = \int \frac{c o s ^{\frac{3}{2}} x}{s i n ^{3} x c o s ^{3} x ( s i n x c o s θ - c o s x s i n θ )} d x$

Manipulating $I_{1}$ for Tangent Form

In $I_{1}$ , factor out $cos x$ from the parenthesis:
$(sin x cos θ - cos x sin θ) = cos x (tan x cos θ - sin θ)$
The denominator of $I_{1}$ becomes:
$sin^{3} x cos^{3} x \cdot cos x (tan x cos θ - sin θ) = sin^{3} x cos^{4} x (tan x cos θ - sin θ)$
$= sin^{\frac{3}{2}} x cos^{2} x tan x cos θ - sin θ$

Simplifying $I_{1}$ to Standard Form

Substitute the simplified denominator back into $I_{1}$ :
$I_{1} = \int \frac{s i n ^{\frac{3}{2}} x}{s i n ^{\frac{3}{2}} x c o s ^{2} x t a n x c o s θ - s i n θ} d x$
$I_{1} = \int \frac{1}{c o s ^{2} x t a n x c o s θ - s i n θ} d x = \int \frac{s e c ^{2} x}{t a n x c o s θ - s i n θ} d x$

Manipulating $I_{2}$ for Cotangent Form

In $I_{2}$ , factor out $sin x$ from the parenthesis:
$(sin x cos θ - cos x sin θ) = sin x (cos θ - cot x sin θ)$
The denominator of $I_{2}$ becomes:
$sin^{3} x cos^{3} x \cdot sin x (cos θ - cot x sin θ) = sin^{4} x cos^{3} x (cos θ - cot x sin θ)$
$= sin^{2} x cos^{\frac{3}{2}} x cos θ - cot x sin θ$

Simplifying $I_{2}$ to Standard Form

Substitute the simplified denominator back into $I_{2}$ :
$I_{2} = \int \frac{c o s ^{\frac{3}{2}} x}{s i n ^{2} x c o s ^{\frac{3}{2}} x c o s θ - c o t x s i n θ} d x$
$I_{2} = \int \frac{1}{s i n ^{2} x c o s θ - c o t x s i n θ} d x = \int \frac{c s c ^{2} x}{c o s θ - c o t x s i n θ} d x$

Solving $I_{1}$ using Substitution

For $I_{1}$ , let $tan x cos θ - sin θ = t^{2}$
Differentiating: $sec^{2} x cos θ d x = 2 t d t ⟹ sec^{2} x d x = \frac{2 t d t}{c o s θ}$
$I_{1} = \int \frac{2 t d t}{c o s θ \cdot t} = \frac{2}{c o s θ} \int d t = \frac{2 t}{c o s θ}$
$I_{1} = 2 sec θ tan x cos θ - sin θ$

Solving $I_{2}$ using Substitution

For $I_{2}$ , let $cos θ - cot x sin θ = z^{2}$
Differentiating: $csc^{2} x sin θ d x = 2 z d z ⟹ csc^{2} x d x = \frac{2 z d z}{s i n θ}$
$I_{2} = \int \frac{2 z d z}{s i n θ \cdot z} = \frac{2}{s i n θ} \int d z = \frac{2 z}{s i n θ}$
$I_{2} = 2 csc θ cos θ - cot x sin θ$

Identifying Coefficients $A$ and $B$

Total integral $I = 2 sec θ tan x cos θ - sin θ + 2 csc θ cos θ - cot x sin θ + C$
Comparing with $A cos θ tan x - sin θ + B cos θ - sin θ cot x + C$ :
$A = 2 sec θ$
$B = 2 csc θ$

Calculating the Product $A B$

Calculate $A B$ :
$A B = (2 sec θ) (2 csc θ)$
$A B = \frac{4}{c o s θ s i n θ}$
Multiply numerator and denominator by $2$ :
$A B = \frac{4 \cdot 2}{2 s i n θ c o s θ} = \frac{8}{s i n 2 θ}$
$A B = 8 csc 2 θ$

Final Conclusion and Key Takeaways

Final Answer: $A B = 8 csc 2 θ$ (Option 4)
Key Takeaway: Strategic factoring within the square root is essential to simplify irrational trigonometric integrals.
Next Challenge: Try solving the same integral if the numerator was $sin^{\frac{3}{2}} x - cos^{\frac{3}{2}} x$ . How would the coefficients change?

00:00 / 00:00

The Sigma Insight: Integration by Substitution

The Art of Trigonometric Decomposition

Imagine you are standing before a massive, intimidating mountain of an integral. At first glance, the expression

\int \frac{sin ^{3/2} x + cos ^{3/2} x}{sin ^{3} x cos ^{3} x sin ( x - θ )} d x

looks like a labyrinth. It is designed to make you panic. But in the world of JEE Advanced, we do not panic; we decompose. We look for the hidden symmetry that the problem setter has carefully tucked away.

Phase 1

The Strategic Split

The first step is to recognize the target. The problem asks us to reach a form involving

A cos θ tan x - sin θ + B cos θ - sin θ cot x + C

. This is our North Star.

It tells us that the integral is not a single monolith, but a sum of two distinct parts. We use the compound angle identity

sin (x - θ) = sin x cos θ - cos x sin θ

to break the denominator open.

By splitting the integral into

I_{1}

and

I_{2}

, we allow ourselves to tackle the

sin^{3/2} x

and

cos^{3/2} x

terms separately.

Phase 2

The Magic of Factoring

Now, let us look at

I_{1}

. We have

sin^{3/2} x

in the numerator. To simplify the denominator, we need to force a

sec^{2} x

to appear.

By factoring

cos x

out of the compound angle term, we get:

sin x cos θ - cos x sin θ = cos x (tan x cos θ - sin θ)

When this is placed inside the square root, the

cos^{3} x

already present in the denominator combines with the new

cos x

to become

cos^{4} x

. When it emerges from the square root, it becomes

cos^{2} x

Suddenly, the integral transforms into the beautiful, manageable form:

I_{1} = \int \frac{sec ^{2} x}{tan x cos θ - sin θ} d x

Phase 3

The Elegance of Substitution

This is where the math starts to sing. We let

t^{2} = tan x cos θ - sin θ

. Differentiating both sides gives us

sec^{2} x cos θ d x = 2 t d t

The

sec^{2} x

term, which we fought so hard to create, cancels out perfectly with the differential. We are left with:

I_{1} = \frac{2}{cos θ} \int d t = 2 sec θ tan x cos θ - sin θ

We repeat this exact logic for

I_{2}

, but this time we factor out

sin x

to create a

csc^{2} x

term. The symmetry is breathtaking. The second part yields:

I_{2} = 2 csc θ cos θ - cot x sin θ

Phase 4

The Final Synthesis

We have arrived at the finish line. By comparing our result with the target form, we identify

A = 2 sec θ

and

B = 2 csc θ

The final task is to calculate the product

A B = (2 sec θ) (2 csc θ)

. Using the identity

sin 2 θ = 2 sin θ cos θ

, we find that:

A B = \frac{4}{sin θ cos θ} = \frac{8}{2 sin θ cos θ} = 8 csc 2 θ

Take a moment to appreciate this. We started with a terrifying radical expression and, through the simple act of factoring and substitution, reduced it to a clean, elegant coefficient. This is the beauty of calculus—no matter how complex the problem, there is always a path to simplicity if you are willing to look for the symmetry.

Common Traps & FAQs

Wait, why do we split the integral into two parts? Can't we just solve it as one big block?

That is a great question! While technically possible, the structure of the target form

A \dots + B \dots

acts as a roadmap. By splitting the integral, we align the numerator terms (

sin^{3/2} x

and

cos^{3/2} x

) with the specific trigonometric identities that simplify the denominator. It turns a chaotic expression into two elegant, solvable pieces.

How did you know to factor out

cos x

in the first part and

sin x

in the second?

It comes down to the goal of creating a derivative. In the first part, we want to use

tan x

, and since the derivative of

tan x

sec^{2} x

, we need to manipulate the denominator to produce

cos^{2} x

outside the radical. Factoring out

cos x

from the compound angle term is the algebraic 'key' that unlocks this transformation.

Why do we use

t^{2}

instead of just

t

for the substitution?

Using

t^{2}

is a classic trick for integrals involving square roots. By setting the entire radical equal to

t

, you get

t = \dots

, which is fine, but setting the *inside* of the radical to

t^{2}

allows the square root to vanish immediately, leaving you with a simple linear integral. It saves you from dealing with messy fractional powers during the integration process!

Is it possible to get a negative sign during the substitution?

Absolutely! Trigonometric functions are periodic and change signs. However, in the context of these indefinite integrals, we are looking for the functional form. If a negative sign appears, it simply gets absorbed into the constant of integration or the coefficients

A

and

B

. Always keep your focus on the derivative matching the numerator.

Why is the final answer

8 csc 2 θ

and not just

4 csc 2 θ

Don't forget the constants! When you differentiate

2 sec θ \dots

, the chain rule brings out a factor of

1/ cos θ

. When you combine the two parts, you get

2 sec θ

and

2 csc θ

. Multiplying these gives

4 sec θ csc θ

, which is

4/ (sin θ cos θ)

. Multiplying by

2/2

gives

8/ (2 sin θ cos θ)

, which is

8 csc 2 θ

Animated Solution for Mathematics - Indefinite Integration: If $\int \frac{s i n ^{3/2} x + c o s ^{3/2} x}{s i n ^{3} x c o s ^{3} x s i n ( x - θ )} d x = A cos θ tan x - sin θ + B cos θ - sin θ cot x + C$ where C is the integration constant, then AB is equal to

Select Answer:

Visualized Solution

Analyzing the Integral Structure

Expanding $sin (x - θ)$

Splitting the Integral into $I_{1}$ and $I_{2}$

Manipulating $I_{1}$ for Tangent Form

Simplifying $I_{1}$ to Standard Form

Manipulating $I_{2}$ for Cotangent Form

Simplifying $I_{2}$ to Standard Form

Solving $I_{1}$ using Substitution

Solving $I_{2}$ using Substitution

Identifying Coefficients $A$ and $B$

Calculating the Product $A B$

Final Conclusion and Key Takeaways

The Sigma Insight: Integration by Substitution

The Art of Trigonometric Decomposition

Phase 1

Phase 2

Phase 3

Phase 4

Similar Questions

The value of the integral $\int \frac{s i n θ - s i n 2 θ ( s i n ^{3} θ + s i n ^{2} θ + s i n θ )}{2 s i n ^{4} θ + 3 s i n ^{2} θ + 6} \frac{d θ}{1 - c o s 2 θ}$ is (where $c$ is a constant of integration)

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 :

The integral $\int sec^{2/3} x csc^{4/3} x d x$ is equal to (Hence C is a constant of integration)

$\int \frac{d x}{c o s x + 3 s i n x}$ equals

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

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 $K$ )

The value of $2 \int \frac{s i n x d x}{s i n ( x - \frac{π}{4} )}$ is

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

The integral $\int \frac{s i n ^{2} x c o s ^{2} x}{( s i n ^{5} x + c o s ^{3} x s i n ^{2} x + s i n ^{3} x c o s ^{2} x + c o s ^{5} x ) ^{2}} d x$ is equal to :

Select Answer:

Visualized Solution

Analyzing the Integral Structure

Expanding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } sin(x−θ)

Splitting the Integral into .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I1​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I2​

Manipulating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I1​ for Tangent Form

Simplifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I1​ to Standard Form

Manipulating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I2​ for Cotangent Form

Simplifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I2​ to Standard Form

Solving .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I1​ using Substitution

Solving .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I2​ using Substitution

Identifying Coefficients .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B

Calculating the Product .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AB

Final Conclusion and Key Takeaways

The Sigma Insight: Integration by Substitution

The Art of Trigonometric Decomposition

Phase 1

Phase 2

Phase 3

Phase 4

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of the integral ∫2sin4θ+3sin2θ+6​sinθ−sin2θ(sin3θ+sin2θ+sinθ)​1−cos2θdθ​ is (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; } The integral ∫sec2/3xcsc4/3xdx is equal to (Hence C is a constant of integration)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∫cosx+3​sinxdx​ equals

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If ∫sin3x(1+sin6x)2/3cosx​dx=f(x)(1+sin6x)1/3+c, where c is a constant of integration, then λf(3π​) is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If ∫sin2x(1+sinx)2/3cosxdx​=f(x)(1+sin6x)1/λ+c where c is a constant of integration, then λf(3π​) is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The integral ∫(secx+tanx)9/2sec2x​dx equals (for some arbitrary constant K)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of 2​∫sin(x−4π​)sinxdx​ is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If ∫sin−1(1+xx​​)dx=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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The integral ∫(sin5x+cos3xsin2x+sin3xcos2x+cos5x)2sin2xcos2x​dx is equal to :

Expanding $sin (x - θ)$

Splitting the Integral into $I_{1}$ and $I_{2}$

Manipulating $I_{1}$ for Tangent Form

Simplifying $I_{1}$ to Standard Form

Manipulating $I_{2}$ for Cotangent Form

Simplifying $I_{2}$ to Standard Form

Solving $I_{1}$ using Substitution

Solving $I_{2}$ using Substitution

Identifying Coefficients $A$ and $B$

Calculating the Product $A B$

The value of the integral $\int \frac{s i n θ - s i n 2 θ ( s i n ^{3} θ + s i n ^{2} θ + s i n θ )}{2 s i n ^{4} θ + 3 s i n ^{2} θ + 6} \frac{d θ}{1 - c o s 2 θ}$ is (where $c$ is a constant of integration)

The integral $\int sec^{2/3} x csc^{4/3} x d x$ is equal to (Hence C is a constant of integration)

$\int \frac{d x}{c o s x + 3 s i n x}$ equals

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

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 $K$ )

The value of $2 \int \frac{s i n x d x}{s i n ( x - \frac{π}{4} )}$ is

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

The integral $\int \frac{s i n ^{2} x c o s ^{2} x}{( s i n ^{5} x + c o s ^{3} x s i n ^{2} x + s i n ^{3} x c o s ^{2} x + c o s ^{5} x ) ^{2}} d x$ is equal to :