JEE Main 2026 (22 January Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Indefinite Integration: If $\int (sin x)^{- 11/2} (cos x)^{- 5/2} d x = - \frac{p _{1}}{q _{1}} (cot x)^{9/2} - \frac{p _{2}}{q _{2}} (cot x)^{5/2} - \frac{p _{3}}{q _{3}} (cot x)^{1/2} + \frac{p _{4}}{q _{4}} (cot x)^{- 3/2} + C$ , where $p_{i}$ and $q_{i}$ are positive integers with $g cd (p_{i}, q_{i}) = 1$ for $i = 1, 2, 3, 4$ and $C$ is the constant of integration, then $\frac{15 p _{1} p _{2} p _{3} p _{4}}{q _{1} q _{2} q _{3} q _{4}}$ is equal to

Enter Numerical Value:

Visualized Solution

Analyze the Integral Form

Given integral: $I = \int (sin x)^{- 11/2} (cos x)^{- 5/2} d x$
Identify powers: $m = - \frac{11}{2}$ and $n = - \frac{5}{2}$
Check sum: $m + n = - \frac{11}{2} - \frac{5}{2} = - \frac{16}{2} = - 8$
Since $m + n$ is a negative even integer, use substitution $t = cot x$ or $t = tan x$ .

Define the Substitution $t = cot x$

Let $t = cot x$
Differentiate: $d t = - csc^{2} x d x$
Express $d x$ : $d x = - \frac{d t}{c s c ^{2} x} = - \frac{d t}{1 + c o t ^{2} x} = - \frac{d t}{1 + t ^{2}}$

Express $sin x$ and $cos x$ in terms of $t$

Using $t = cot x$ :
$sin x = \frac{1}{1 + t ^{2}} = (1 + t^{2})^{- 1/2}$
$cos x = \frac{t}{1 + t ^{2}} = t (1 + t^{2})^{- 1/2}$

Substitute into the Integral

Substitute $sin x$ , $cos x$ , and $d x$ :
$I = \int [(1 + t^{2})^{- 1/2}]^{- 11/2} \cdot [t (1 + t^{2})^{- 1/2}]^{- 5/2} \cdot (- \frac{d t}{1 + t ^{2}})$

Simplify the Algebraic Expression

Combine powers of $(1 + t^{2})$ :
Power $= \frac{11}{4} + \frac{5}{4} - 1 = \frac{16}{4} - 1 = 4 - 1 = 3$
Simplified Integral: $I = - \int t^{- 5/2} (1 + t^{2})^{3} d t$

Expand the Binomial $(1 + t^{2})^{3}$

Expand using $(a + b)^{3} = a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3}$ :
$(1 + t^{2})^{3} = 1 + 3 t^{2} + 3 t^{4} + t^{6}$

Distribute $t^{- 5/2}$

Distribute $t^{- 5/2}$ into the expansion:
$I = - \int (t^{- 5/2} + 3 t^{- 1/2} + 3 t^{3/2} + t^{7/2}) d t$

Integrate Term by Term

Apply $\int t^{n} d t = \frac{t ^{n + 1}}{n + 1}$ :
$I = - [\frac{t ^{- 3/2}}{- 3/2} + 3 \frac{t ^{1/2}}{1/2} + 3 \frac{t ^{5/2}}{5/2} + \frac{t ^{9/2}}{9/2}] + C$

Simplify the Coefficients

Simplify each fraction:
$I = - [- \frac{2}{3} t^{- 3/2} + 6 t^{1/2} + \frac{6}{5} t^{5/2} + \frac{2}{9} t^{9/2}] + C$

Distribute the Negative Sign

Distribute the negative sign:
$I = - \frac{2}{9} t^{9/2} - \frac{6}{5} t^{5/2} - 6 t^{1/2} + \frac{2}{3} t^{- 3/2} + C$

Identify $p_{i}$ and $q_{i}$ by Comparison

Compare with $- \frac{p _{1}}{q _{1}} t^{9/2} - \frac{p _{2}}{q _{2}} t^{5/2} - \frac{p _{3}}{q _{3}} t^{1/2} + \frac{p _{4}}{q _{4}} t^{- 3/2} + C$ :
$\frac{p _{1}}{q _{1}} = \frac{2}{9} ⟹ p_{1} = 2, q_{1} = 9$
$\frac{p _{2}}{q _{2}} = \frac{6}{5} ⟹ p_{2} = 6, q_{2} = 5$
$\frac{p _{3}}{q _{3}} = \frac{6}{1} ⟹ p_{3} = 6, q_{3} = 1$
$\frac{p _{4}}{q _{4}} = \frac{2}{3} ⟹ p_{4} = 2, q_{4} = 3$

Final Calculation

Calculate: $\frac{15 p _{1} p _{2} p _{3} p _{4}}{q _{1} q _{2} q _{3} q _{4}}$
$= 15 \cdot \frac{2 \cdot 6 \cdot 6 \cdot 2}{9 \cdot 5 \cdot 1 \cdot 3}$
$= 15 \cdot \frac{144}{135} = \frac{2160}{135} = 16$
Final Answer: 16

00:00 / 00:00

The Sigma Insight: Evaluation of Special Integral Forms

Analyzing the Setup

The integral to evaluate is:

I = \int (sin x)^{- 11/2} (cos x)^{- 5/2} d x

To solve this, we perform a diagnostic on the exponents

m = - 11/2

and

n = - 5/2

. Their sum is

m + n = - 8

, which is a negative even integer.

This specific property indicates that the integral can be simplified significantly by using the substitution

t = cot x

The Transformation

We set

t = cot x

, which implies

d t = - csc^{2} x d x

. Consequently, the differential becomes:

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

Using the geometric relationship where

cot x = t

, we identify the trigonometric functions in terms of

t

sin x = (1 + t^{2})^{- 1/2}, cos x = t (1 + t^{2})^{- 1/2}

The Algebraic Cleanup

Substituting these expressions into the original integral, we obtain:

I = \int [(1 + t^{2})^{- 1/2}]^{- 11/2} \cdot [t (1 + t^{2})^{- 1/2}]^{- 5/2} \cdot (- \frac{d t}{1 + t ^{2}})

Combining the powers of

(1 + t^{2})

, the integral simplifies to:

I = - \int t^{- 5/2} (1 + t^{2})^{3} d t

Expanding the polynomial

(1 + t^{2})^{3} = 1 + 3 t^{2} + 3 t^{4} + t^{6}

, we distribute the

t^{- 5/2}

term:

I = - \int (t^{- 5/2} + 3 t^{- 1/2} + 3 t^{3/2} + t^{7/2}) d t

The Final Integration

Applying the power rule

\int t^{n} d t = \frac{t ^{n + 1}}{n + 1}

, we integrate term by term:

I = - [\frac{t ^{- 3/2}}{- 3/2} + 3 \frac{t ^{1/2}}{1/2} + 3 \frac{t ^{5/2}}{5/2} + \frac{t ^{9/2}}{9/2}] + C

Simplifying the coefficients, we arrive at the final expression:

I = - \frac{2}{9} t^{9/2} - \frac{6}{5} t^{5/2} - 6 t^{1/2} + \frac{2}{3} t^{- 3/2} + C

By identifying the constants

p_{1} = 2, q_{1} = 9, p_{2} = 6, q_{2} = 5, p_{3} = 6, q_{3} = 1, p_{4} = 2, q_{4} = 3

, we perform the final calculation:

\frac{15 p _{1} p _{2} p _{3} p _{4}}{q _{1} q _{2} q _{3} q _{4}} = 15 \cdot \frac{2 \cdot 6 \cdot 6 \cdot 2}{9 \cdot 5 \cdot 1 \cdot 3} = * * 16 * *

Common Traps & FAQs

Wait, why do we choose

t = cot x

instead of

t = tan x

? Does it even matter?

That is a fantastic question! In this specific problem, the final expression is given in terms of

cot x

, so choosing

t = cot x

aligns perfectly with the target format, saving you from having to convert back at the end. Mathematically, both

t = tan x

and

t = cot x

would work because the sum of the exponents is a negative even integer, but choosing the one that matches your final answer format is a strategic move that saves time and prevents algebraic errors.

I got lost in the algebra when simplifying

(1 + t^{2})^{3}

. Is there a trick to avoid expanding it?

I understand the temptation to look for a shortcut! However, in this case, the expansion is actually quite friendly. Since the power is only 3, the binomial expansion

(1 + t^{2})^{3} = 1 + 3 t^{2} + 3 t^{4} + t^{6}

is straightforward. Trying to use more complex methods like integration by parts or reduction formulas would likely be overkill and much more prone to error. Embrace the algebra—sometimes the most direct path is the most reliable one.

Why did the power of

(1 + t^{2})

simplify so perfectly to 3? Is that just luck?

It is not luck at all—it is the beauty of the problem design! The exponents were specifically chosen so that when you convert

sin x

and

cos x

into terms of

t

, the powers of

(1 + t^{2})

combine to form an integer. This is the hallmark of a well-crafted JEE Advanced problem. It rewards you for identifying the correct substitution by 'collapsing' the complexity into a simple polynomial.

Enter Numerical Value:

Visualized Solution

Analyze the Integral Form

Define the Substitution $t = cot x$

Express $sin x$ and $cos x$ in terms of $t$

Substitute into the Integral

Simplify the Algebraic Expression

Expand the Binomial $(1 + t^{2})^{3}$

Distribute $t^{- 5/2}$

Integrate Term by Term

Simplify the Coefficients

Distribute the Negative Sign

Identify $p_{i}$ and $q_{i}$ by Comparison

Final Calculation

The Sigma Insight: Evaluation of Special Integral Forms

Analyzing the Setup

The Transformation

The Algebraic Cleanup

The Final Integration

Similar Questions

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Evaluate $\int \frac{s i n x}{s i n x - c o s x} d x$

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Enter Numerical Value:

Visualized Solution

Analyze the Integral Form

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

Substitute into the Integral

Simplify the Algebraic Expression

Expand the Binomial .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (1+t2)3

Distribute .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t−5/2

Integrate Term by Term

Simplify the Coefficients

Distribute the Negative Sign

Identify .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } pi​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } qi​ by Comparison

Final Calculation

The Sigma Insight: Evaluation of Special Integral Forms

Analyzing the Setup

The Transformation

The Algebraic Cleanup

The Final Integration

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