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Some JEE integrals have limits like 0 to 10π or 0 to 50π, and trying to integrate directly means splitting at every point where the function changes sign. If the function is periodic there is a shortcut: work out one period and multiply. This post explains why the shortcut works, how to use it when the limits are not a whole number of periods, and where it breaks.
Where is this in the syllabus?
The JEE (Main) 2026 syllabus lists "properties of definite integrals" and their evaluation in the integral calculus unit, and the JEE (Advanced) 2026 brochure lists "definite integral and their properties". The CBSE 2026-27 curriculum asks for basic properties of definite integrals in Class 12. The NCERT Class 12 chapter Integrals (2026-27 reprint) states properties P0 to P7, but the periodic-function property is not among them. It follows from P2, so you can prove it with NCERT tools, as below. Check the current documents for your exam.
Why is the integral over n periods n times one period?
Suppose f(x + T) = f(x) for every x. NCERT's P2 lets you split an integral at any point, so split 0 to nT into n pieces:
∫ from 0 to nT = ∫ from 0 to T + ∫ from T to 2T + … + ∫ from (n − 1)T to nT
In the piece from kT to (k + 1)T, substitute x = t + kT. Then dx = dt, the limits become 0 and T, and f(t + kT) = f(t) because f repeats. So every piece equals ∫ from 0 to T of f(t) dt, and there are n of them:
∫ from 0 to nT of f(x) dx = n ∫ from 0 to T of f(x) dx, for n a positive integer.
The same idea gives a second, often more useful rule. The integral over any interval of length T is the same:
∫ from a to a + T of f(x) dx = ∫ from 0 to T of f(x) dx, for every a.
Think of a repeating wallpaper: any strip one pattern wide contains the same total amount of ink.
Worked example 1: |sin x| from 0 to 10π
|sin x| has period π, because sin(x + π) = −sin x and the modulus removes the sign. Over one period:
∫ from 0 to π of sin x dx = [−cos x] from 0 to π = 1 + 1 = 2.
From 0 to 10π there are 10 periods, so the integral is 10 × 2 = 20. A numerical check by computer gives 20.000 to three decimal places.
Note that ∫ from 0 to 10π of sin x dx, without the modulus, is 0, because each positive arch is cancelled by a negative one. The modulus changes the period and the answer.
Worked example 2: sin² x from 0 to 50π
sin² x = (1 − cos 2x)/2 has period π. Over one period:
∫ from 0 to π of (1 − cos 2x)/2 dx = π/2 − 0 = π/2.
From 0 to 50π there are 50 periods, so the integral is 50 × π/2 = 25π.
This is the same fact that sits behind root mean square values in physics. The NCERT Class 12 Physics chapter Alternating Current uses the average of sin² ωt over a complete cycle, which is 1/2, to find the average power in a resistor. Integrating over a whole number of cycles and dividing by the time gives the same average.
Worked example 3: when the limits are not whole periods
Find ∫ from 0 to 7π/2 of |sin x| dx.
7π/2 = 3π + π/2: three full periods and half of one more.
Three periods: 3 × 2 = 6.
The leftover piece, 3π to 7π/2, is by the shifting rule the same as 0 to π/2: ∫ from 0 to π/2 of sin x dx = 1.
Total: 6 + 1 = 7.
And a shifted window: ∫ from π/4 to π/4 + 6π of |sin x| dx. The interval has length 6π, six periods, so the starting point does not matter. The answer is 6 × 2 = 12.
Worked example 4: rewrite to find the period
Find ∫ from 0 to 20π of |sin x + cos x| dx.
Write sin x + cos x = √2 sin(x + π/4). The modulus of a sine has period π, so one period contributes √2 × 2 = 2√2. From 0 to 20π there are 20 periods, so the answer is 20 × 2√2 = 40√2, about 56.57. Writing the function in one-sine form is the step that reveals the period. Our post on periodic, even and odd functions has more on finding periods.
Which mistakes cost marks?
Using a number that is not a period. sin x does not repeat after π, so ∫ from 0 to 10π of sin x dx is not 10 × 2 = 20; it is 0. Any true period works, even one that is not the smallest.
Applying the rule to a non-periodic function. x + sin x does not repeat, so ∫ from 0 to 10π is not 10 times anything.
Forgetting the leftover piece when the interval is not a whole number of periods.
Mixing up this rule with P4. P4 replaces x by a − x and says nothing about repetition.
A practice routine
Prove the n-periods rule yourself from P2 and the substitution x = t + kT, without looking.
Evaluate ∫ from 0 to 8π of |cos x| dx and ∫ from 0 to 6π of cos² x dx.
Do one problem with a leftover piece, such as |cos x| from 0 to 9π/2.
Then combine it with the King's property in our post on properties of definite integrals.
If the substitution step is where you get stuck, the Fermi Concept Library covers every concept in mathematics with its own lesson, so you can revise substitution in definite integrals first.
Today, write the proof of the n-periods rule in five lines and check it against the one above.
