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JEE Mains Maths · Definite Integration

Reduction Formulas and Beta Integrals

Families of integrals indexed by n, linked by a relation between neighbouring members, and the Beta integral B(m, n) with its symmetry and its sum rule.

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Why this matters

Twelve PYQs. None asks for a single integral's value: each asks for a relation between members of a family, or a ratio, sum or combination of them, which the reduction formula collapses. Two ideas cover the page.

Concept 1 of 2: Reduction formulas by parts

Integrate InI_n by parts, or split off one power, to express it through In−1I_{n-1} or In−2I_{n-2}. Once the relation is known, ratios like InIn+1\frac{I_n}{I_{n+1}} and sums like In+In+2I_n+I_{n+2} collapse to simple expressions in nn.

Definition

  • ∫01(1−xk)n dx=In\int_0^1(1-x^k)^n\,dx=I_n: In=nknk+1In−1I_n=\frac{nk}{nk+1}I_{n-1}.
  • ∫0π/4tan⁡nx dx=In\int_0^{\pi/4}\tan^nx\,dx=I_n: In+In+2=1n+1I_n+I_{n+2}=\frac1{n+1}.
  • ∫0π/2sin⁡nx dx=In\int_0^{\pi/2}\sin^nx\,dx=I_n: In=n−1nIn−2I_n=\frac{n-1}nI_{n-2}.
  • ∫01xnex dx=In\int_0^1x^ne^x\,dx=I_n: In=e−nIn−1I_n=e-nI_{n-1}.

Powers of sine

∫0π/2sin⁡nx dx=n−1n∫0π/2sin⁡n−2x dx\int_0^{\pi/2}\sin^n x\,dx=\frac{n-1}{n}\int_0^{\pi/2}\sin^{n-2}x\,dx

Worked example

In=∫01xnex dxI_n=\int_0^1x^ne^x\,dx. Find I2I_2.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2021 · Paper 4 · Q64Moderate

Example 1 · Definite Integration · Reduction Formulas and Beta Integrals

If In=∫π/4π/2cot⁡nxdxI_{n}=\int_{\pi/4}^{\pi/2} \cot^{n}xdx, then:

Keep the boundary term

The term [uv][uv] from integrating by parts is often what gives the relation its constant, as the ee in In=e−nIn−1I_n=e-nI_{n-1}. Dropping it makes the relation wrong.
Drill 8 more on reduction formulas by parts

Concept 2 of 2: Beta integrals

B(m,n)=∫01xm−1(1−x)n−1dxB(m,n)=\int_0^1x^{m-1}(1-x)^{n-1}dx. Putting x→1−xx\to1-x shows B(m,n)=B(n,m)B(m,n)=B(n,m). Since x+(1−x)=1x+(1-x)=1, the integrand of B(m,n)B(m,n) splits into those of B(m+1,n)B(m+1,n) and B(m,n+1)B(m,n+1). For whole numbers, B(m,n)=(m−1)! (n−1)!(m+n−1)!B(m,n)=\frac{(m-1)!\,(n-1)!}{(m+n-1)!}. The substitution t=xkt=x^k turns ∫01(1−xk)n dx\int_0^1(1-x^k)^n\,dx into a Beta integral.

Definition

  • B(m,n)=∫01xm−1(1−x)n−1dx=B(n,m)B(m,n)=\int_0^1x^{m-1}(1-x)^{n-1}dx=B(n,m).
  • B(m+1,n)+B(m,n+1)=B(m,n)B(m+1,n)+B(m,n+1)=B(m,n).
  • Whole numbers: B(m,n)=(m−1)!(n−1)!(m+n−1)!B(m,n)=\frac{(m-1)!(n-1)!}{(m+n-1)!}.
  • ∫01(1−xk)n dx=1kB(1k,n+1)\int_0^1(1-x^k)^n\,dx=\frac1kB\left(\frac1k,n+1\right).

The sum rule

B(m+1,n)+B(m,n+1)=B(m,n)B(m+1,n)+B(m,n+1)=B(m,n)

Worked example

Evaluate ∫01x(1−x)3 dx\int_0^1x(1-x)^3\,dx.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 24 Jan 2025 · Q52Moderate

Example 2 · Definite Integration · Reduction Formulas and Beta Integrals

In I(m,n)=∫01xm−1(1−x)n−1dx,m,n>0I(m,n) =\int_{0}^{1} x^{m - 1}(1 - x)^{n - 1}dx,m,n > 0, then I(9,14)+I(10,13)I(9,14) + I(10,13) is

The exponents are m − 1 and n − 1

x2(1−x)2x^2(1-x)^2 is B(3,3)B(3,3), not B(2,2)B(2,2). Add 1 to each exponent to read off mm and nn.
Drill 2 more on beta integrals

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

  • Reduction formulas by parts

    Powers of sine

    ∫0π/2sin⁡nx dx=n−1n∫0π/2sin⁡n−2x dx\int_0^{\pi/2}\sin^n x\,dx=\frac{n-1}{n}\int_0^{\pi/2}\sin^{n-2}x\,dx
  • Beta integrals

    The sum rule

    B(m+1,n)+B(m,n+1)=B(m,n)B(m+1,n)+B(m,n+1)=B(m,n)

Watch out for (2)

Test yourself on Definite Integration

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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