Definite integrals by substitution
NCERT Class 12 Mathematics Chapter 7 Section 7.9
Quick answer
To evaluate a definite integral by substitution, put t = g(x) and change the limits too: x = a and x = b become t = g(a) and t = g(b). The integral is then worked out entirely in t, without returning to x. Keep the new limits in their order even when the lower one is larger, carry the sign of dt, and use only a substitution that is defined and continuous on the whole interval.
With your textbook
NCERT Mathematics Part II (Class 12), Chapter 7 “Integrals”, Section 7.9 “Evaluation of Definite Integrals by Substitution”, pages 271–273.
Open the official NCERT chapter (PDF, 5.7 MB, ncert.nic.in) (opens in a new tab)
Checked against the 2026-27 reprint on .
How is a definite integral found by substitution?
Substitution in a definite integral
Here, the integral will be kept in the new variable itself, and the limits of the integral will accordingly be changed, so that we can perform the last step.
NCERT Mathematics Part II (Class 12), Section 7.9, page 272
- NCERT’s steps (page 271): substitute y = f(x) or x = g(y) in the integral without limits; integrate in the new variable; put the old variable back; then take the difference of the values at the upper and lower limits.
- NCERT’s Note (page 272) shortens this: change the limits to the new variable and skip putting the old one back.
In plain words: a substitution changes the variable, so the limits must change with it. If t = g(x), then x = a becomes t = g(a) and x = b becomes t = g(b). Once the limits are in t, the answer never needs to be written in x again.
Changing the limits
NCERT Example 26 (page 272): for ∫−11 5x4√(x5 + 1) dx, put t = x5 + 1, so dt = 5x4 dx. When x = −1, t = 0; when x = 1, t = 2. So
∫−11 5x4√(x5 + 1) dx = ∫02 √t dt
= (2/3)[t3/2] from 0 to 2 = (2/3)(2√2) = 4√2/3.
NCERT also works it the long way: find (2/3)(x5 + 1)3/2 first, then put in x = 1 and x = −1. Both give 4√2/3.
NCERT Example 27 (pages 272–273): for ∫01 tan x/(1 + x2) dx, put t = tan x. The limits 0 and 1 become 0 and π/4, the principal values of tan 0 and tan 1, and
∫0π/4 t dt = ½(π/4)2 = π2/32.
The order of the new limits matters. If t = cos x on [0, π/2], then x = 0 gives t = 1 and x = π/2 gives t = 0, so the new lower limit is larger than the new upper one. Keep them in that order and carry the minus sign of dt = −sin x dx. Swapping the limits later (NCERT P, page 273) changes the sign again, which cancels it.
Solved examples
Example 1: A log from a substitution (NCERT Exercise 7.9, Q1, page 273)
Evaluate ∫01 x/(x2 + 1) dx.
Put t = x2 + 1, so x dx = ½ dt; the limits become 1 and 2. The integral is ½∫12 dt/t = ½(log 2 − log 1).
Answer: ½ log 2.
Example 2: Powers of sin and cos (NCERT Exercise 7.9, Q2, page 273)
Evaluate ∫0π/2 √(sin φ) cos5 φ dφ.
Put t = sin φ, so dt = cos φ dφ and cos4 φ = (1 − t2)2; the limits become 0 and 1. The integral is ∫01 (t1/2 − 2t5/2 + t9/2) dt = 2/3 − 4/7 + 2/11.
Answer: 64/231.
Example 3: Limits that swap (NCERT Exercise 7.9, Q5, page 273)
Evaluate ∫0π/2 sin x/(1 + cos2 x) dx.
Put t = cos x, so sin x dx = −dt; x = 0 gives t = 1 and x = π/2 gives t = 0. The integral is −∫10 dt/(1 + t2) = ∫01 dt/(1 + t2) = tan 1.
Answer: π/4.
Example 4: Complete the square (NCERT Exercise 7.9, Q7, page 273)
Evaluate ∫−11 dx/(x2 + 2x + 5).
x2 + 2x + 5 = (x + 1)2 + 22. Put t = x + 1; the limits become 0 and 2. The integral is ½[tan (t/2)] from 0 to 2 = ½(π/4 − 0).
Answer: π/8.
Example 5: Choose the correct answer (NCERT Exercise 7.9, Q9, page 273)
The value of ∫1/31 (x − x3)1/3/x4 dx is (A) 6; (B) 0; (C) 3; (D) 4.
Take x3 out of the cube root: (x − x3)1/3 = x(1/x2 − 1)1/3, so the integrand is (1/x2 − 1)1/3/x3. Put t = 1/x2 − 1, so dt = −2 dx/x3; x = 1/3 gives t = 8 and x = 1 gives t = 0. The integral is ½∫08 t1/3 dt = ½ × (3/4) × 84/3 = ½ × (3/4) × 16 = 6.
Answer: 6, (A).
NCERT slip (answers, page 446): the key gives “9. D”, that is 4. The value is 6, option (A): the substitution above gives it exactly, and a numerical check agrees. In an exam, trust the working and choose (A).
Common mistakes to avoid
Common mistake
Keeping the old limits after substituting
In NCERT Example 26, ∫√t dt from −1 to 1 is wrong: −1 and 1 are values of x. With t = x5 + 1 the limits are 0 and 2.
Common mistake
Reordering the new limits
With t = cos x on [0, π/2], the limits in t are 1 (lower) and 0 (upper). Swapping them to 0 and 1 without also dropping the minus sign of dt = −sin x dx changes the sign of the answer.
Common mistake
Substituting with a function that breaks on the interval
Beyond the NCERT text (JEE): x = 1/t in ∫−11 dx/(1 + x2) seems to give −π/2, but the value is π/2 (the integrand is positive). 1/t is not defined at t = 0, between the new limits −1 and 1, so this substitution is not allowed on that interval.
Quick revision summary
- In ∫ab f(x) dx, a substitution t = g(x) changes the limits: x = a and x = b become t = g(a) and t = g(b); then integrate in t and stop.
- Keep the new limits in order, lower first, even when the lower one is larger, and carry the sign of dt.
- The substitution must be defined and continuous on the whole interval.
Formula sheet
- t = g(x): limits a, b → g(a), g(b)
- ∫ f(g(x)) g′(x) dx from a to b = ∫ f(t) dt from g(a) to g(b)
Questions students ask
What is u-substitution in a definite integral, and why do the limits change?
The u-substitution of many textbooks is NCERT’s substitution with u for t; in a definite integral the limits change because, for continuous f and g′, ∫ab f(g(x)) g′(x) dx and the integral of f(u) from g(a) to g(b) both equal F(g(b)) − F(g(a)), where F′ = f, by the chain rule and the second theorem. So the new limits are the values of u at the old ones. For ∫01 3x2(x3 + 1)4 dx, u = x3 + 1 runs from 1 to 2, giving ∫12 u4 du = 31/5.
How do you choose the new limits in a trigonometric substitution?
In a trigonometric substitution, take each new limit as a principal value of the inverse function, which keeps the substitution one-one and fixes the sign of the square root: with x = 2 sin θ, θ = sin (x/2) lies in [−π/2, π/2], where cos θ ≥ 0, so √(4 − x2) is 2 cos θ. For ∫01 x2/√(4 − x2) dx, x = 1 gives sin θ = ½ and θ = π/6, not 5π/6, and the integral becomes ∫0π/6 4 sin2 θ dθ, which is π/3 − √3/2, about 0.181. NCERT takes principal values in the same way for tan in its Example 27 (pages 272–273).
Does the substitution have to be one-one on the interval of integration?
A substitution t = g(x) need not be one-one when the integrand is exactly f(g(x)) g′(x) with f and g′ continuous (reasoned from the second fundamental theorem): t = sin x is not one-one on [0, π], yet ∫0π sin2 x cos x dx becomes ∫00 t2 dt = 0, which is right. It must be one-one, or the interval split, when x has to be written back in terms of t: ∫0π 1 dx is π, but t = sin x makes both limits 0, because dx = dt/cos x and cos x is √(1 − t2) on [0, π/2] but −√(1 − t2) on [π/2, π]. Split at π/2, the two halves give π/2 each.
Practice questions
Try each question first, then open the answer to check your working.
-
Q1. Evaluate ∫01 sin (2x/(1 + x2)) dx (NCERT Exercise 7.9, Q3, page 273).
Show answerHide answer
With x = tan θ, 2x/(1 + x2) = sin 2θ, and for 0 ≤ x ≤ 1, 2θ is in [0, π/2], so the integrand is 2 tan x. By parts, ∫2 tan x dx = 2x tan x − log (1 + x2), which from 0 to 1 gives π/2 − log 2.
-
Q2. Evaluate ∫02 x√(x + 2) dx, putting x + 2 = t2 (NCERT Exercise 7.9, Q4, page 273).
Show answerHide answer
dx = 2t dt, and the limits become √2 and 2. The integral is ∫2t2(t2 − 2) dt = [2t5/5 − 4t3/3] from √2 to 2 = 32/15 + 16√2/15 = 16√2(√2 + 1)/15.
-
Q3. Evaluate ∫12 (1/x − 1/(2x2)) e2x dx (NCERT Exercise 7.9, Q8, page 273).
Show answerHide answer
Put t = 2x: the limits become 2 and 4, and the integral is ∫24 (1/t − 1/t2)et dt, of the form et[f + f′] with f = 1/t. So it is [et/t] from 2 to 4 = e4/4 − e2/2 = e2(e2 − 2)/4.
-
Q4. If f(x) = ∫0x t sin t dt, then f′(x) is (A) cos x + x sin x; (B) x sin x; (C) x cos x; (D) sin x + x cos x (NCERT Exercise 7.9, Q10, page 273).
Show answerHide answer
By the first fundamental theorem, f′(x) = x sin x: (B).