The chain rule in reverse: master u-substitution to evaluate complex integrals.
The substitution method (also called u-substitution or change of variables) is the integration counterpart to the chain rule for differentiation. If we know that:
Then reversing this:
If is a differentiable function and is continuous, then:
where we substitute back after integration.
Problem: Evaluate
Solution:
Let , then
Problem: Evaluate
Solution:
Let , then
Problem: Evaluate
Solution:
Let , then , so
→ Let u = ax + b
→ Let u = f(x)
→ Let u = f(x)
→ Let u = f(x)
Problem: Evaluate
Solution:
Recognize: numerator = derivative of denominator
Problem: Derive
Solution:
Write as . Let u = cos x, du = -sin x dx
| Expression | Substitution | Identity Used |
|---|---|---|
Problem: Evaluate
Solution:
Let , then
Problem: Evaluate
Solution:
Let , then
Back-substitute:
Look for a function and its derivative together. Common patterns: f(g(x))·g'(x), composite functions, expressions under radicals.
Often you can adjust by constants. If you need 2x dx but have x dx, multiply inside and divide outside by 2.
For √(a²-x²) use x = a sin θ. For √(a²+x²) use x = a tan θ. For √(x²-a²) use x = a sec θ.
Forgetting to change dx to du, or not substituting back to x at the end.
Yes! Some integrals require successive substitutions. Just keep track of all variables.
If you see a composite function f(g(x)) and g'(x) (or a constant multiple) is present, try substitution.
Always replace dx with du using du = g'(x) dx.
The final answer must be in terms of x, not u.
Usually u should be the “inner” function whose derivative appears.
Problem:
Let u = x³, du = 3x² dx, so x² dx = du/3:
Problem:
Let u = ln x, du = dx/x:
Problem:
Let u = sin x, du = cos x dx:
Problem:
Let u = x²+1, du = 2x dx:
Problem:
Let u = sec x, du = sec x tan x dx:
Problem:
Let u = arctan x, du = dx/(1+x²):
Problem:
Let x = 3 sin θ, dx = 3 cos θ dθ:
Back-substitute: , ,
Problem:
Let x = 2 tan θ, dx = 2 sec²θ dθ:
,
Use tan²θ = sec²θ - 1:
Problem: (x > 3)
Let x = 3 sec θ, dx = 3 sec θ tan θ dθ:
Back:
1.
2.
Let u = √x:
3.
4.
Let u = eˣ:
5.
Let u = x⁴:
6.
Let u = tan x:
Problem:
Let u = ln x, then du = dx/x. Let v = √u, then dv = du/(2√u):
Problem:
Complete: x²+4x+8 = (x+2)² + 4. Let u = x+2:
Problem:
Let u = √x, so x = u², dx = 2u du:
Problem:
Let x = sinh t, dx = cosh t dt:
The derivative of the inner function should appear (perhaps with a constant).
Inner function, exponent, argument of trig/log, expression under radical.
If du has extra constants, factor them out or multiply/divide as needed.
√(a²-x²) → x=a sin θ, √(a²+x²) → x=a tan θ, √(x²-a²) → x=a sec θ
For quadratics in denominator, complete square then substitute.
Final answer must be in terms of original variable.
| Integrand Pattern | Substitution | Result Form |
|---|---|---|
| u = ax+b | ||
| u = g(x) | ||
| u = g(x) | ||
| x = a sin θ | Trig result | |
| x = a tan θ | Trig result | |
| x = a sec θ | Trig result |
Challenge 1:
u = 1+x³:
Challenge 2:
u = 1+eˣ:
Challenge 3:
u = cos x:
Challenge 4:
u = x²:
Problem:
Let u = tan x, du = sec²x dx:
Problem:
Let u = x³, du = 3x² dx:
Problem:
Let u = √x, du = dx/(2√x):
Problem:
Let u = x²-1, du = 2x dx:
Problem:
Let u = arctan x, du = dx/(1+x²):
P1:
P2:
P3:
P4:
P5:
P6:
| When You See | Try This u | Because du = |
|---|---|---|
| u = x^n | nx^{n-1}dx | |
| u = e^x | e^x dx | |
| u = ln x | dx/x | |
| u = sin x | cos x dx | |
| u = cos x | -sin x dx | |
| u = tan x | sec² x dx | |
| u = sec x | sec x tan x dx | |
| u = arctan x | dx/(1+x²) |
Problem:
Let x = 2 sin θ:
Problem:
Let x = 3 tan θ:
Problem: (x > 4)
Let x = 4 sec θ:
Result:
Problem:
Complete: x²-6x+13 = (x-3)² + 4
Let u = x-3:
Problem:
Complete: -x²+4x+5 = 9-(x-2)²
For ∫f(g(x))g'(x)dx, let u = g(x). The integral becomes ∫f(u)du.
For expressions with √(a²-x²), √(a²+x²), or √(x²-a²), use appropriate trig substitution.
For quadratic expressions, complete the square first, then substitute.
Always differentiate your answer to verify it's correct.
P7:
P8:
P9:
P10:
C5:
Let u = x^(1/6), then x = u⁶
C6:
Try x = cos θ or rationalize
C7:
u = x² gives du = 2x dx
Problem:
Let u = √x, so x = u², dx = 2u du:
Problem:
Let u = √x, du = dx/(2√x):
Problem:
Let x = sinh t, dx = cosh t dt:
Problem:
Let √(x²+1) - x = t, then x = (1-t²)/(2t):
x = sin θ
u = eˣ+1
u = x² gives du = 2x dx
| Expression | Substitution |
|---|---|
| √(a²-x²) | x = a sin θ |
| √(a²+x²) | x = a tan θ or x = a sinh t |
| √(x²-a²) | x = a sec θ or x = a cosh t |
| f(g(x))·g'(x) | u = g(x) |
| √(ax²+bx+c) | Complete square first |
Identify inner function u and check if du is present
For √(a²±x²), √(x²-a²) expressions
For ax²+bx+c before substitution
Inner function + its derivative (or constant multiple)
For √(a²-x²), √(a²+x²), √(x²-a²)
Differentiate your answer to check