01 · Map the bounds
For u = x² + 3 and x from 1 to 3, what are the u bounds?
Hint
Substitute each x endpoint separately.
Worked solution
4 to 12.
Understand · explore · practise
Definite integrals by substitution
Evaluate definite integrals by substitution. Change both bounds, track negative differentials and reversed limits, preserve exact answers and check the interval and branch.
Before you startIntegration by substitution, definite integrals and exact values.
01 / Transform both endpoints
Bounds belong to the variable you are integrating.
∫ₐᵇ F(g(x))g′(x) dx = ∫g(a)g(b) F(u) du
Use u = g(x) and check that the functions are valid throughout the interval.
When the integral changes from x to u, the endpoints must change too. Put each original x endpoint into u = g(x). Do not keep x bounds on a u integral. The order of the transformed endpoints carries the orientation.
Follow the endpoints into uExplore
x: 0 → 2 becomes u: 1 → 5.
Original integrand: 2x(x² + 1)². Transformed integrand: u².
The signed integral is [(5)³ − (1)³]/3 = 41.333333.
The u bounds run upwards.
These examples integrate g′(x)[g(x)]² dx, so the entire differential becomes u² du. The two number lines use different scales.
02 / Build an endpoint table
Transform the integrand and bounds as separate steps.
Evaluate ∫₀² x√(x² + 5) dx.Worked example
u = x² + 5; x dx = du/2
The transformed integrand is (1/2)√u.
x = 0 → u = 5; x = 2 → u = 9
Transform both endpoints.
(1/2)∫₅⁹ u¹ᐟ² du = [u³ᐟ²/3]₅⁹
Integrate entirely in u.
9 − (5√5)/3
Keep the answer exact.
Watch: decreasing substitutions reverse the endpoints
Pause, replay or seek freely. The notes explain the same idea and stay in view.
03 / Evaluate using the new bounds
There is no need to return to x if the bounds are already in u.
Evaluate ∫₀¹ 6x(x² + 2)² dx.Worked example
u = x² + 2; 6x dx = 3du
The factor 3 remains.
x = 0 → u = 2; x = 1 → u = 3
The transformed bounds are 2 and 3.
3∫₂³ u² du = [u³]₂³
Evaluate the u primitive.
27 − 8 = 19
A definite integral has a numerical value.
02 · A logarithmic definite integral
Evaluate ∫₀² 2x/(x² + 1) dx using u = x² + 1.
Hint
The u bounds are 1 and 5.
Worked solution
∫₁⁵ du/u = ln 5.
03 · No surviving constant
Why is there no + C in the final value of a definite integral?
Hint
Subtract the primitive’s endpoint values.
Worked solution
The same constant appears at both endpoints and cancels.
04 / Keep a negative differential
A decreasing substitution produces descending u bounds.
Evaluate ∫₀² (5 − 2x)² dx.Worked example
u = 5 − 2x; dx = −du/2
The differential has a negative factor.
x = 0 → u = 5; x = 2 → u = 1
Do not reorder the bounds silently.
−(1/2)∫₅¹ u² du
Both the negative factor and descending bounds are correct.
(1/2)∫₁⁵ u² du = 62/3
Reversing the bounds removes the minus sign.
04 · Preserve orientation
With u = 7 − x, rewrite ∫₁³ 1/(7 − x) dx.
Hint
The bounds change from 6 to 4 and dx = −du.
Worked solution
−∫₆⁴ du/u = ∫₄⁶ du/u = ln(3/2).
05 · Spot a double reversal
A student changes −∫₅¹ u² du into −∫₁⁵ u² du. What is wrong?
Hint
Swapping bounds changes the sign.
Worked solution
The correct expression is +∫₁⁵ u² du. Keeping the minus sign reverses the value incorrectly.
05 / Transform trigonometric endpoints exactly
Calculate the new endpoint values before integrating.
Evaluate ∫₀^(π/2) cos x/(2 + sin x) dx.Worked example
u = 2 + sin x; du = cos x dx
Use radians.
x = 0 → u = 2; x = π/2 → u = 3
Use exact sine values.
∫₂³ du/u = ln(3/2)
The denominator stays positive throughout.
06 · Cosine reverses the bounds
Evaluate ∫₀^(π/2) sin x/(3 + cos x)² dx.
Hint
Use u = 3 + cos x, from 4 to 3, and du = −sin x dx.
Worked solution
−∫₄³ u⁻² du = [u⁻¹]₄³ = 1/3 − 1/4 = 1/12.
07 · A shifted upper endpoint
If u = sin x and x runs from 0 to π/6, what are the u bounds?
Hint
sin(π/6) is an exact value.
Worked solution
0 to 1/2.
06 / Respect a root substitution’s range
u = √x chooses the nonnegative branch.
Evaluate ∫₁⁴ 1/(1 + √x) dx.Worked example
u = √x; dx = 2u du
The u bounds are 1 and 2.
∫₁² [2 − 2/(1 + u)] du
Simplify 2u/(1 + u).
[2u − 2ln(1 + u)]₁²
Use u values, not the original x bounds.
2 − 2ln(3/2)
The result is positive.
08 · Root endpoints
With u = √(x + 1), map x = 3 and x = 8.
Hint
Use the principal nonnegative square root.
Worked solution
u = 2 and u = 3.
09 · A root cancellation
Evaluate ∫₁⁹ 1/[√x(1 + √x)] dx.
Hint
u = √x gives 2∫₁³ du/(1 + u).
Worked solution
2ln 4 − 2ln 2 = 2ln 2.
07 / Use exact exponential bounds
A logarithmic x endpoint can turn into a simple u value.
Evaluate ∫₀^(ln 3) e²ˣ/(1 + eˣ) dx.Worked example
u = eˣ; dx = du/u
The integrand becomes u/(1 + u).
x = 0 → u = 1; x = ln 3 → u = 3
Use e^(ln 3) = 3.
∫₁³ [1 − 1/(1 + u)] du
Divide first.
[u − ln(1 + u)]₁³ = 2 − ln 2
Combine the logarithms exactly.
10 · Exponential bounds
For u = e²ˣ, map x = 0 and x = ln 2.
Hint
e^(2ln 2) = 4.
Worked solution
1 to 4.
08 / You can substitute back instead
Use one complete route at a time.
Evaluate ∫₀² x√(x² + 5) dx by returning to x.Worked example
A primitive in u is u³ᐟ²/3
This follows from u = x² + 5.
Return to (x² + 5)³ᐟ²/3
Now the expression is in x again.
Use the original x bounds 0 and 2
[ (x² + 5)³ᐟ²/3 ]₀² = 9 − 5√5/3.
Both routes are valid: evaluate a u primitive at u bounds, or an x primitive at x bounds. Mixing the two is the error.
11 · Detect mixed bounds
After using u = x² + 5 for x from 0 to 2, a student evaluates [u³ᐟ²/3]₀². Repair the bounds.
Hint
Those are x values on a u expression.
Worked solution
Use [u³ᐟ²/3]₅⁹, or substitute back and then use 0 and 2.
12 · Reversed original limits
If ∫₀² x√(x² + 5) dx = 9 − 5√5/3, what is ∫₂⁰ x√(x² + 5) dx?
Hint
Reversing the original limits also changes the sign.
Worked solution
5√5/3 − 9.
09 / Check the entire interval
A change of variable does not remove a singularity.
For ∫₀² 1/(x − 1) dx, the denominator vanishes at x = 1. Changing to u = x − 1 sends the interval to −1 through 0 to 1; the singularity is still there. Endpoint logarithms do not give an ordinary finite integral.
If you solve for x using a square root or inverse trigonometric function, make sure the chosen branch represents the original interval. A substitution that is not one-to-one may need separate interval pieces when you use its inverse. Direct chain-rule integrals can sometimes be checked without taking an inverse.
13 · Branch warning
With u = x² and x from −2 to −1, is x = √u the correct inverse?
Hint
The original x values are negative.
Worked solution
No. Use x = −√u on that interval. The u endpoints run from 4 to 1.
10 / Check the sign and scale
Use simple bounds to catch an implausible result.
If the original integrand is positive and the original lower bound is smaller than the upper bound, the definite integral must be positive. A decreasing u substitution does not change that fact; its negative differential and reversed u limits work together.
Keep radicals, logarithms and multiples of π exact until a decimal is requested. A quick numerical estimate is a check, not a replacement for the exact calculation.
14 · Equal limits
What is ∫₁¹ 2x/(x² + 3) dx?
Hint
The interval has zero width.
Worked solution
0. The transformed bounds are also equal: 4 to 4.
15 · A sign test
A student gets −62/3 for ∫₀² (5 − 2x)² dx. Why must that be wrong?
Hint
The squared integrand is positive on [0,2].
Worked solution
The integral must be positive. The correct value is 62/3; a sign was lost in the decreasing substitution.
11 / Keep bounds, variable and differential together
Write a short endpoint table before evaluating.
- State u and transform the whole differential.
- Map each original endpoint through u = g(x).
- Preserve the endpoint order or explicitly reverse the sign.
- Check the interval and any inverse branch.
- Evaluate in one variable and keep exact values.
16 · Complete a new calculation
Evaluate ∫₀¹ 4x/(x² + 2)² dx.
Hint
u = x² + 2 gives 2∫₂³ u⁻² du.
Worked solution
[−2/u]₂³ = −2/3 + 1 = 1/3.
Section 1 of 11 · Transform both endpoints