The chain rule differentiates composite functions: the derivative of sin(x2) is cos(x2)⋅2x — differentiate the outer function, then multiply by the derivative of the inside. Integration asks the reverse: given cos(x2)⋅2x, find the original function. That's what u substitution does — it reverses the chain rule.
The Core Intuition
When an integral looks like "a function times the derivative of its inside," substitute the inside with u and the derivative of the inside with du. Consider:
∫2xcos(x2)dx
Here 2x is the derivative of x2, and x2 is the inside of cos(x2). Let u=x2, so du=2xdx:
∫cos(u)du=sin(u)+C=sin(x2)+C
Check: the derivative of sin(x2) is cos(x2)⋅2x.
The Precise Statement
∫f(g(x))⋅g′(x)dx=∫f(u)duwhere u=g(x),du=g′(x)dx
Valid provided g is differentiable and the resulting integral in u is simpler.
The Step-by-Step Method
Identify a function g(x) whose derivative g′(x) also appears (possibly up to a constant factor).
Setu=g(x), compute du=g′(x)dx.
Rewrite the entire integral in u and du — every x and dx must be replaced.
Integrate with respect to u.
Substitute backu=g(x).
Watch out
You cannot mix variables. If any x remains after substitution, you chose the wrong u (or must solve for x in terms of u — rare).
A Second Example (with a constant factor)
Evaluate ∫xx2+1dx. Let u=x2+1, so xdx=21du:
∫u⋅21du=21⋅32u3/2+C=31(x2+1)3/2+C
When Does It Work?
When the integrand is something times the derivative of something inside. Common patterns:
The numerator equals exactly 21 of the derivative of the denominator, so the integral is 21log∣6cosx+4sinx∣+C.
Spot the pattern
Whenever the top of a fraction is (a constant times) the derivative of the bottom, the integral is a logarithm: ∫g(x)g′(x)dx=log∣g(x)∣+C. So differentiate the denominator and compare.
Check the relationship
Denominator g(x)=6cosx+4sinx, so
g′(x)=−6sinx+4cosx=4cosx−6sinx=2(2cosx−3sinx).
That is exactly twice the numerator, so the numerator is 21g′(x) — with no left-over constant term. (If you set numerator =Ag+Bg′ and match the coefficients of cosx and sinx, you get A=0, B=21.)
Method: Numerator = constant multiple of the denominator's derivative
Use this for a trig ratio like 6cosx+4sinx2cosx−3sinx where the top turns out to be a constant times the derivative of the bottom — the integral is then a logarithm.
Steps
Step 1: Differentiate the denominator and compare.
dxd(6cosx+4sinx)=−6sinx+4cosx=2(2cosx−3sinx), exactly twice the numerator.
Mistake 1: Not checking whether the numerator is the denominator's derivative.
Why it's wrong: dxd(6cosx+4sinx)=4cosx−6sinx=2(2cosx−3sinx); missing this leaves the integral looking impossible. Correct approach: differentiate the denominator and compare with the numerator.