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JEE Mains Maths · Definite Integration

Odd, Even and Periodic Integrands

Integrals over an interval symmetric about 0, where odd parts vanish and even parts double; the 1/(1 + aᵍ⁽ˣ⁾) denominator that halves an even integrand; and integrals of periodic functions over many periods.

Why this matters

Twenty-five PYQs, most with limits −a to a. The integrand usually has an odd part that drops out, or a denominator like 1 + eˣ that pairs with its reflection to give 1. A few integrate a periodic function over many periods. Three ideas cover the page.

Concept 1 of 3: Odd parts vanish, even parts double

On [−a,a][-a,a] an odd integrand contributes 0 and an even one contributes twice its integral on [0,a][0,a]. So split the integrand into its odd and even parts before doing anything else. A term like x3x2+2∣x∣+1\frac{x^3}{x^2+2|x|+1} is odd, because everything but x3x^3 depends only on ∣x∣|x|.

Definition

  • ff odd: ∫−aaf=0\int_{-a}^af=0. ff even: ∫−aaf=2∫0af\int_{-a}^af=2\int_0^af.
  • odd × even = odd; odd × odd = even.
  • ∣x∣|x|, x2x^2, cos⁡x\cos x are even; xx, sin⁡x\sin x, ln⁡(x+x2+1)\ln\big(x+\sqrt{x^2+1}\big) are odd.

Symmetric limits

∫−aaf={0,f odd2∫0af,f even\int_{-a}^{a}f=\begin{cases}0,&f\text{ odd}\\2\int_0^af,&f\text{ even}\end{cases}

Worked example

Evaluate ∫−11(x3+x2+x) dx\int_{-1}^1(x^3+x^2+x)\,dx.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 6 Apr 2026 Shift 2 · Q70Moderate

Example 1 · Definite Integration · Odd, Even and Periodic Integrands

The value of the integral ∫−11(x3+∣x∣+1x2+2∣x∣+1)dx\int_{- 1}^{1} \left( \frac{x^{3}+ |x| + 1}{x^{2}+ 2|x| + 1} \right)dx is equal to

Check the whole term

x∣x∣x|x| is odd but x2∣x∣x^2|x| is even. Decide the parity of each complete term, not of its pieces.

Concept 2 of 3: A denominator 1 + bᵍ⁽ˣ⁾ with g odd

If gg is odd and hh is even, replacing xx by −x-x turns h(x)1+bg(x)\frac{h(x)}{1+b^{g(x)}} into h(x) bg(x)1+bg(x)\frac{h(x)\,b^{g(x)}}{1+b^{g(x)}}. The two add to h(x)h(x), so the integral over [−a,a][-a,a] is half of ∫−aah\int_{-a}^ah, which is ∫0ah\int_0^ah.

Definition

  • ∫−aah(x)1+bg(x)dx=∫0ah(x) dx\int_{-a}^a\frac{h(x)}{1+b^{g(x)}}dx=\int_0^ah(x)\,dx (hh even, gg odd, b>0b>0).
  • Common gg: xx, sin⁡x\sin x, x∣x∣x|x|, xcos⁡xx\cos x.
  • exex+e−x=11+e−2x\frac{e^x}{e^x+e^{-x}}=\frac1{1+e^{-2x}} is the same form.

Halving an even integrand

∫−aah(x)1+bg(x)dx=∫0ah(x) dx\int_{-a}^{a}\frac{h(x)}{1+b^{g(x)}}dx=\int_0^{a}h(x)\,dx

Worked example

Evaluate ∫−11x21+2xdx\int_{-1}^1\frac{x^2}{1+2^x}dx.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2021 · Paper 25 · Q75Moderate

Example 2 · Definite Integration · Odd, Even and Periodic Integrands

The value of ∫−π/2π/2cos⁡2x1+3xdx\int_{-\pi/2}^{\pi/2} \frac{\cos^{2}x}{1 +3^{x}}dx is:

The numerator must be even

An odd numerator over 1+bg1+b^{g} does not halve. Split the numerator into even and odd parts and treat each by its own rule.

Concept 3 of 3: Periodic integrands

If ff repeats every TT, every interval of length TT gives the same integral, so ∫0nTf=n∫0Tf\int_0^{nT}f=n\int_0^Tf. A relation like f(x)+f(x+k)=cf(x)+f(x+k)=c makes ff repeat every 2k2k, and each stretch of length 2k2k integrates to ckck.

Definition

  • f(x+T)=f(x)⇒∫aa+nTf=n∫0Tff(x+T)=f(x)\Rightarrow\int_a^{a+nT}f=n\int_0^Tf.
  • Periods: ∣sin⁡x∣|\sin x|, sin⁡2x\sin^2x, sin⁡4x+cos⁡4x\sin^4x+\cos^4x: π\pi (the last is even π2\frac\pi2).
  • f(x)+f(x+k)=c⇒f(x)+f(x+k)=c\Rightarrow period 2k2k and ∫aa+2kf=ck\int_a^{a+2k}f=ck.

Many periods

∫0nTf(x) dx=n∫0Tf(x) dx\int_0^{nT}f(x)\,dx=n\int_0^{T}f(x)\,dx

Worked example

Evaluate ∫010π∣sin⁡x∣ dx\int_0^{10\pi}|\sin x|\,dx.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 2 Apr 2026 Shift 2 · Q67Moderate

Example 3 · Definite Integration · Odd, Even and Periodic Integrands

The value of ∫020π(sin⁡4x+cos⁡4x)dx\int_{0}^{20\pi} \left( \sin^{4}x +\cos^{4}x \right)dx is equal to:

The right period

∣sin⁡x∣|\sin x| repeats every π\pi, not 2π2\pi. Using the longer period halves the number of copies and halves the answer.

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 (3)

  • Odd parts vanish, even parts double

    Symmetric limits

    ∫−aaf={0,f odd2∫0af,f even\int_{-a}^{a}f=\begin{cases}0,&f\text{ odd}\\2\int_0^af,&f\text{ even}\end{cases}
  • A denominator 1 + bᵍ⁽ˣ⁾ with g odd

    Halving an even integrand

    ∫−aah(x)1+bg(x)dx=∫0ah(x) dx\int_{-a}^{a}\frac{h(x)}{1+b^{g(x)}}dx=\int_0^{a}h(x)\,dx
  • Periodic integrands

    Many periods

    ∫0nTf(x) dx=n∫0Tf(x) dx\int_0^{nT}f(x)\,dx=n\int_0^{T}f(x)\,dx

Watch out for (3)

Test yourself on Definite Integration

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