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Integration By Substitution

Integration by substitution reverses the chain rule by transforming complicated integrals into simpler forms. Identify an inner function u=g(x)u = g(x), replace g′(x) dxg'(x)\,dx with dudu, integrate with respect to uu, then back-substitute to recover the answer in terms of xx.

Integration by Substitution

Integration by substitution is the reverse of the chain rule. It transforms integrals of the form ∫f(g(x)) g′(x) dx\int f(g(x))\,g'(x)\,dx into simpler ones by letting u=g(x)u = g(x), so du=g′(x) dxdu = g'(x)\,dx, yielding ∫f(u) du.\int f(u)\,du.

Choosing Your Substitution

The key is spotting the inner function — typically something nested inside a power, root, trigonometric, exponential, or logarithmic function. Check whether its derivative (or a constant multiple of it) appears elsewhere in the integrand. If it does, set uu equal to that inner function, compute the differential dudu, and rewrite the entire integral in terms of uu only. No xx variables should remain.

Indefinite Integrals

After integrating with respect to uu, back-substitute u=g(x)u = g(x) to express your answer in the original variable.

Example: For ∫2xcos⁡(x2) dx\int 2x\cos(x^2)\,dx, let u=x2u = x^2, so du=2x dxdu = 2x\,dx: ∫cos⁡(u) du=sin⁡(u)+C=sin⁡(x2)+C.\int \cos(u)\,du = \sin(u) + C = \sin(x^2) + C.

Definite Integrals

For ∫abf(g(x))g′(x) dx\int_a^b f(g(x))g'(x)\,dx, transform the bounds directly: when x=ax=a, u=g(a)u=g(a); when x=bx=b, u=g(b)u=g(b). Then evaluate entirely in uu, avoiding the need to back-substitute.

Example: For ∫02xex2 dx\int_0^2 x e^{x^2}\,dx, let u=x2u = x^2, so x dx=12dux\,dx = \frac{1}{2}du. Bounds: x=0⇒u=0x=0 \Rightarrow u=0; x=2⇒u=4x=2 \Rightarrow u=4: 12∫04eu du=12(e4−1).\frac{1}{2}\int_0^4 e^u\,du = \frac{1}{2}(e^4 - 1).

Practical Note

Substitution simplifies integrals arising in applications like the Net Change Theorem. When velocity involves composite functions, substitution finds the antiderivative needed to compute displacement or distance.

StepAction
1. IdentifyFind the inner function g(x)g(x)
2. DifferentiateCompute du=g′(x) dxdu = g'(x)\,dx
3. SubstituteReplace all xx-terms with uu-terms; update bounds if definite
4. IntegrateEvaluate ∫f(u) du\int f(u)\,du in terms of uu
5. Back-substituteReplace uu with g(x)g(x) (indefinite) or evaluate at new bounds (definite)

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