JEE MainMathematics

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Combining Beta Function Integrals via Algebraic Factorization

Q: Is the trigonometric substitution necessary, or can we directly combine the integrals in terms of x?

Direct combination in x is even faster: x^8(1-x)^13 + x^9(1-x)^12 = x^8(1-x)^12[(1-x) + x] = x^8(1-x)^12.

In I(m,n)=integral from 0 to 1 of x^(m-1)(1-x)^(n-1)dx, m, n>0, then I(9,14)+I(10,13) is

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AI(9,1)BI(19,27)CI(1,13)DI(9,13)

Step-by-step solutionView

Correct answer

By combining the integrals into a single integrand and factoring out $x^{8} (1 - x)^{12}$ , the sum simplifies directly to $I (9, 13)$ .

Option analysis

Why each option works or fails

A · I(9,1)

Believing that subtracting indices or canceling exponents yields $I (9, 14 - 13) = I (9, 1)$ . Factor out common powers from the integrands: $x^{8} (1 - x)^{12} [(1 - x) + x] = x^{8} (1 - x)^{12}$ , which corresponds to $I (9, 13)$ , not $I (9, 1)$ .

B · I(19,27)

Believing that the integral operator adds the arguments directly as $I (9 + 10, 14 + 13) = I (19, 27)$ . Integrals cannot be combined by adding their parameters; write out the integrands explicitly and simplify algebraically.

C · I(1,13)

Incorrectly simplifying the powers to leave only $(1 - x)^{12}$ with power 0 on $x$ , leading to $I (1, 13)$ . Ensure you identify the common factor correctly: both terms share $x^{8}$ , not $x^{0}$ , giving parameters $m = 9$ and $n = 13$ .

D · I(9,13)

This is the correct option. Factoring the combined integrand gives $\int_{0}^{1} x^{8} (1 - x)^{12} (1 - x + x) d x = \int_{0}^{1} x^{9 - 1} (1 - x)^{13 - 1} d x = I (9, 13)$ .

Reviewed route

Solution

StepWorking

01Given

$I (m, n) = \int_{0}^{1} x^{m - 1} (1 - x)^{n - 1} d x$ for $m, n > 0$ . We are required to find $I (9, 14) + I (10, 13)$ .

02Approach

Substitute $x = \sin^{2} θ$ , so $d x = 2 \sin θ \cos θ d θ$ . This converts $I (m, n)$ into $2 \int_{0}^{π / 2} (\sin θ)^{2 m - 1} (\cos θ)^{2 n - 1} d θ$ . Then combine the integrals using $\sin^{2} θ + \cos^{2} θ = 1$ .

03Execute

For $I (9, 14)$ and $I (10, 13)$ : $I (9, 14) = 2 \int_{0}^{π / 2} (\sin θ)^{17} (\cos θ)^{27} d θ$ $I (10, 13) = 2 \int_{0}^{π / 2} (\sin θ)^{19} (\cos θ)^{25} d θ$ Summing them: $I (9, 14) + I (10, 13) = 2 \int_{0}^{π / 2} (\sin θ)^{17} (\cos θ)^{25} [\cos^{2} θ + \sin^{2} θ] d θ$ $= 2 \int_{0}^{π / 2} (\sin θ)^{17} (\cos θ)^{25} d θ$ Since $2 m - 1 = 17 ⟹ m = 9$ and $2 n - 1 = 25 ⟹ n = 13$ , this is $I (9, 13)$ .

✓Verify

Check via algebraic property of Beta function: $I (m, n + 1) + I (m + 1, n) = \int_{0}^{1} x^{m - 1} (1 - x)^{n - 1} [(1 - x) + x] d x = \int_{0}^{1} x^{m - 1} (1 - x)^{n - 1} d x = I (m, n)$ . Setting $m = 9, n = 13$ gives $I (9, 14) + I (10, 13) = I (9, 13)$ .

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 7 September 2026

Quick checks

Students also ask

Is the trigonometric substitution necessary, or can we directly combine the integrals in terms of x?

Direct combination in x is even faster: $x^{8} (1 - x)^{13} + x^{9} (1 - x)^{12} = x^{8} (1 - x)^{12} [(1 - x) + x] = x^{8} (1 - x)^{12}$ .

Answer

By combining the integrals into a single integrand and factoring out $x^{8} (1 - x)^{12}$ , the sum simplifies directly to $I (9, 13)$ .

Why each option works or fails

A: I(9,1) - Believing that subtracting indices or canceling exponents yields $I (9, 14 - 13) = I (9, 1)$ . Factor out common powers from the integrands: $x^{8} (1 - x)^{12} [(1 - x) + x] = x^{8} (1 - x)^{12}$ , which corresponds to $I (9, 13)$ , not $I (9, 1)$ .
B: I(19,27) - Believing that the integral operator adds the arguments directly as $I (9 + 10, 14 + 13) = I (19, 27)$ . Integrals cannot be combined by adding their parameters; write out the integrands explicitly and simplify algebraically.
C: I(1,13) - Incorrectly simplifying the powers to leave only $(1 - x)^{12}$ with power 0 on $x$ , leading to $I (1, 13)$ . Ensure you identify the common factor correctly: both terms share $x^{8}$ , not $x^{0}$ , giving parameters $m = 9$ and $n = 13$ .
D · correct: I(9,13) - This is the correct option. Factoring the combined integrand gives $\int_{0}^{1} x^{8} (1 - x)^{12} (1 - x + x) d x = \int_{0}^{1} x^{9 - 1} (1 - x)^{13 - 1} d x = I (9, 13)$ .

Step-by-step solution

Given: $I (m, n) = \int_{0}^{1} x^{m - 1} (1 - x)^{n - 1} d x$ for $m, n > 0$ . We are required to find $I (9, 14) + I (10, 13)$ .
Approach: Substitute $x = \sin^{2} θ$ , so $d x = 2 \sin θ \cos θ d θ$ . This converts $I (m, n)$ into $2 \int_{0}^{π / 2} (\sin θ)^{2 m - 1} (\cos θ)^{2 n - 1} d θ$ . Then combine the integrals using $\sin^{2} θ + \cos^{2} θ = 1$ .
Execute: For $I (9, 14)$ and $I (10, 13)$ : $I (9, 14) = 2 \int_{0}^{π / 2} (\sin θ)^{17} (\cos θ)^{27} d θ$ $I (10, 13) = 2 \int_{0}^{π / 2} (\sin θ)^{19} (\cos θ)^{25} d θ$ Summing them: $I (9, 14) + I (10, 13) = 2 \int_{0}^{π / 2} (\sin θ)^{17} (\cos θ)^{25} [\cos^{2} θ + \sin^{2} θ] d θ$ $= 2 \int_{0}^{π / 2} (\sin θ)^{17} (\cos θ)^{25} d θ$ Since $2 m - 1 = 17 ⟹ m = 9$ and $2 n - 1 = 25 ⟹ n = 13$ , this is $I (9, 13)$ .
Verify: Check via algebraic property of Beta function: $I (m, n + 1) + I (m + 1, n) = \int_{0}^{1} x^{m - 1} (1 - x)^{n - 1} [(1 - x) + x] d x = \int_{0}^{1} x^{m - 1} (1 - x)^{n - 1} d x = I (m, n)$ . Setting $m = 9, n = 13$ gives $I (9, 14) + I (10, 13) = I (9, 13)$ .

Shortcut:

When to use it: Fastest route: combine integrands directly under a single integral without changing variables.

Given: $I (9, 14) = \int_{0}^{1} x^{8} (1 - x)^{13} d x$ and $I (10, 13) = \int_{0}^{1} x^{9} (1 - x)^{12} d x$ .

Execute: $I (9, 14) + I (10, 13) = \int_{0}^{1} [x^{8} (1 - x)^{13} + x^{9} (1 - x)^{12}] d x = \int_{0}^{1} x^{8} (1 - x)^{12} [(1 - x) + x] d x = \int_{0}^{1} x^{8} (1 - x)^{12} d x = I (9, 13)$

Verify: Since $(1 - x) + x = 1$ , the degree in $x$ is 8 and in $(1 - x)$ is 12, which matches $I (8 + 1, 12 + 1) = I (9, 13)$ .

Combining Beta Function Integrals via Algebraic Factorization

What feels right?

By combining the integrals into a single integrand and factoring out $x^{8} (1 - x)^{12}$ , the sum simplifies directly to $I (9, 13)$ .

Why each option works or fails

Solution

01Given

02Approach

03Execute

✓Verify

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