U-substitution on a definite integral works exactly like it does on an indefinite integral, with one extra step: you have to handle the limits of integration. You can either convert the limits to u-values immediately (cleaner), or back-substitute to get x back before evaluating (familiar). Both work. Most people find Method 1 faster once they have done it a few times.

Two Methods

When applying u-substitution to a definite integral ∫ₐᵇ, you have two choices:

📋 Method Comparison
Method 1Convert bounds: when substituting u = g(x), change limits to u = g(a) and u = g(b). Evaluate without converting back.
Method 2Keep original bounds: find the antiderivative in terms of u, convert back to x, then evaluate at x = a and x = b.

Method 1 is usually faster and less error-prone. Method 2 is useful when you need the antiderivative for other purposes.

Example 1 — Convert the Bounds

📋 ∫₀² 2x(x²+1)³ dx
Set uu = x²+1, du = 2x dx
New boundsx=0 → u=1, x=2 → u=5
Integral∫₁⁵ u³ du = [u⁴/4]₁⁵ = 625/4 − 1/4 = 156

Example 2 — Trig Inside

📋 ∫₀^(π/2) sin(x)cos(x) dx
Set uu = sin(x), du = cos(x) dx
New boundsx=0 → u=0, x=π/2 → u=1
Integral∫₀¹ u du = [u²/2]₀¹ = 1/2

Example 3 — Exponential

📋 ∫₀¹ xe^(x²) dx
Set uu = x², du = 2x dx → x dx = du/2
New boundsx=0 → u=0, x=1 → u=1
Integral½∫₀¹ eᵘ du = ½[eᵘ]₀¹ = ½(e−1) ≈ 0.859

Example 4 — Rational Function

📋 ∫₁³ 2x/(x²+1) dx
Set uu = x²+1, du = 2x dx
New boundsx=1 → u=2, x=3 → u=10
Integral∫₂¹⁰ (1/u) du = [ln|u|]₂¹⁰ = ln10 − ln2 = ln5 ≈ 1.609

Example 5 — Reverse Bounds

📋 ∫₀^(π) cos(x)sin³(x) dx
Set uu = sin(x), du = cos(x) dx
New boundsx=0 → u=0, x=π → u=0
Answer∫₀⁰ u³ du = 0 (upper = lower bound)

Example 6 — Off by a Constant

📋 ∫₁² x²(x³−1)⁴ dx
Set uu = x³−1, du = 3x² dx → x² dx = du/3
New boundsx=1 → u=0, x=2 → u=7
Integral(1/3)∫₀⁷ u⁴ du = (1/3)[u⁵/5]₀⁷ = 7⁵/15 = 16807/15 ≈ 1120.5

Example 7 — Negative du

📋 ∫₀^(π/2) sin(π/2 − x) dx
Set uu = π/2−x, du = −dx → dx = −du
New boundsx=0 → u=π/2, x=π/2 → u=0
Integral−∫_(π/2)⁰ sin(u) du = ∫₀^(π/2) sin(u) du = [−cos u]₀^(π/2) = 1

Example 8 — e^(−x)

📋 ∫₀¹ e^(−2x) dx
Set uu = −2x, du = −2 dx → dx = −du/2
New boundsx=0 → u=0, x=1 → u=−2
Integral−½∫₀^(−2) eᵘ du = ½∫_(−2)⁰ eᵘ du = ½[eᵘ]_(−2)⁰ = ½(1−e^(−2)) ≈ 0.432
⚠ The Most Common Error

Forgetting to change the limits. If you substitute u = g(x) but leave the limits as x = a and x = b, your answer is wrong. Either change both limits or convert back to x before evaluating.

Frequently Asked Questions

What happens if the new lower bound is larger than the upper bound?▾
That is fine — you evaluate normally. ∫ₐᵇ with a > b just gives a negative of ∫ᵦᵃ. You can also flip the bounds and negate the integral if you prefer working with the smaller number on the bottom.
Can I always change the bounds instead of converting back?▾
Yes — for definite integrals, changing bounds is always valid and usually cleaner. It avoids the extra step of substituting x back in. The only case where you might not want to is if you need the antiderivative expression in terms of x for some other purpose.

When the Substitution Flips the Bounds

Sometimes after substituting, the new lower bound is larger than the new upper bound. This is perfectly fine — you evaluate normally. The integral automatically handles the direction. Alternatively, flip the bounds and negate: ∫ₐᵇ = −∫ᵦᵃ.

📋 ∫₁⁰ 2x(1−x²)³ dx — bounds flip
u = 1−x²du = −2x dx → 2x dx = −du
New boundsx=1 → u=0, x=0 → u=1. So u goes from 0 to 1.
Integral∫₀¹ u³·(−du) = −[u⁴/4]₀¹ = −1/4

Trig Substitution in Definite Integrals

📋 ∫₀^(π/4) tan(x)sec²(x) dx
u = tan(x)du = sec²(x) dx
New boundsx=0 → u=0, x=π/4 → u=1
Integral∫₀¹ u du = [u²/2]₀¹ = 1/2

Recognising When u-Sub Works

U-substitution works when the integrand contains a composite function f(g(x)) AND the derivative g'(x) also appears as a factor (possibly multiplied by a constant). The tell-tale sign: one part of the integrand "looks like" the derivative of another part.

📋 Quick recognition drill
∫ x·e^(x²) dx✅ u-sub: x² inside eˣ, and x = (1/2)·d/dx[x²] appears
∫ x·eˣ dx❌ u-sub fails: x is not the derivative of x. Use IBP instead.
∫ sin(x)/cos²(x) dx✅ u = cos(x), du = −sin(x)dx → −∫u⁻² du = 1/cos(x) + C
∫ x/(x²+1) dx✅ u = x²+1, du = 2x dx → (1/2)ln(x²+1) + C
AM
Dr. Aisha Malik, PhD Mathematics
Senior Lecturer in Applied Mathematics · 12 years teaching calculus
Dr. Malik holds a PhD in Applied Mathematics from the University of Edinburgh and has taught calculus to over 4,000 students. She has reviewed this article for mathematical accuracy and pedagogical clarity.
Technically reviewed by: Prof. James Chen, Stanford Mathematics Department · April 2026