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Substitution in a definite integral has a shortcut that saves a line and removes a common slip: convert the limits along with the variable, and never go back to x. Students who learn it often hit two problems. The new limits come out "upside down" and they panic, or they use a substitution that quietly breaks inside the interval. This post covers both, with four worked integrals and two traps.
Where is this in the syllabus?
The JEE (Main) 2026 syllabus lists integration by substitution and the evaluation of definite integrals in Unit 8, and the JEE (Advanced) 2026 brochure lists "integration by the methods of substitution" and definite integrals. The NCERT Class 12 chapter Integrals (reprint 2026-27) has a section on it, 7.9, Evaluation of Definite Integrals by Substitution. Check the current year's documents.
What does NCERT's shortcut say?
Section 7.9 first gives four steps: substitute, integrate, return to x, then put in the limits. Its note then says that "there is no need of step 3": keep the integral in the new variable and change the limits accordingly. Example 27 does this with t = tan⁻¹x, where x running from 0 to 1 becomes t running from 0 to π/4.
The rule is simple. If t = g(x), then x = a becomes t = g(a) and x = b becomes t = g(b). The lower limit of t comes from the lower limit of x, even if it turns out larger.
Worked example 1: the limits flip
Find the integral of sin x cos²x from 0 to π/2.
Put t = cos x, so dt = −sin x dx.
Limits: x = 0 gives t = 1; x = π/2 gives t = 0.
The integral becomes the integral from 1 to 0 of −t² dt.
Swap the limits and drop the minus sign: the integral from 0 to 1 of t² dt = 1/3.
A reversed pair of limits is not an error. Swapping them changes the sign, and here that cancels the minus from dt.
Worked example 2: a trig substitution
Find the integral of √(1 − x²) from 0 to 1.
Put x = sin θ, so dx = cos θ dθ. On 0 ≤ θ ≤ π/2, cos θ ≥ 0, so √(1 − x²) = cos θ.
Limits: x = 0 gives θ = 0; x = 1 gives θ = π/2.
The integral becomes the integral from 0 to π/2 of cos²θ dθ = the integral of (1 + cos 2θ)/2, which is [θ/2 + (sin 2θ)/4] from 0 to π/2 = π/4.
Check. The curve y = √(1 − x²) from 0 to 1 is a quarter of the unit circle, with area π/4. Here the new limits did more than save a step: they told you which sign of cos θ to use.
Worked example 3: leftover x
Find the integral of x/√(x + 1) from 0 to 3.
Put t = √(x + 1), so x = t² − 1 and dx = 2t dt.
Limits: x = 0 gives t = 1; x = 3 gives t = 2.
The integrand becomes (t² − 1)/t × 2t dt = 2(t² − 1) dt.
The integral is 2[t³/3 − t] from 1 to 2 = 2[(8/3 − 2) − (1/3 − 1)] = 2(2/3 + 2/3) = 8/3.
Worked example 4: a log substitution
Find the integral of (ln x)²/x from 1 to e. Put t = ln x, dt = dx/x. Limits: 1 gives 0 and e gives 1. The integral is the integral from 0 to 1 of t² dt = 1/3.
Trap 1: a substitution that is not one-to-one
Consider the integral of x² from −1 to 1, which is 2/3. Put t = x² blindly: both limits become t = 1, and the "answer" is 0. The problem is that t = x² folds the interval at x = 0, so x = √t holds only for x ≥ 0. The fix is to split the integral at 0, or simply not to use that substitution.
Trap 2: a substitution that breaks inside the interval
The integral of 1/(1 + x²) from −1 to 1 is tan⁻¹1 − tan⁻¹(−1) = π/2. Try t = 1/x: then dx = −dt/t², 1 + x² = (t² + 1)/t², and the integral becomes the integral of −1/(1 + t²) dt from −1 to 1, which is −π/2. The sign is wrong because t = 1/x is not defined at x = 0, which lies inside the interval.
The test for both traps is the same: the substitution must be continuous and one-to-one on the whole interval. NCERT's examples all meet this; exam traps are built to break it.
Which mistakes cost marks?
Changing the limits and then also going back to x.
Forgetting the constant factor in dx, such as dt = 2x dx.
Panicking at reversed limits instead of swapping them with a sign change.
Using t = x², t = 1/x or t = tan x on an interval where they fold or break.
The method itself, and how to pick t, is in our post on integration by substitution; for integrals that need a property instead, see properties of definite integrals.
If the area picture behind a definite integral is hard to hold in your head, the Fermi Simulation Lab has interactive simulations for the ideas that are easier to see than to read.
A practice routine
Do NCERT Exercise 7.9 with the shortcut only, writing the new limits on the substitution line.
Redo three of them the long way and compare.
Make one trap of each kind yourself and explain in a sentence why it fails.
Check every answer with a sketch or a quick numerical estimate.
Today, evaluate the integral of x(1 − x²)³ from 0 to 1 with t = 1 − x², and notice which way the limits go.
