Substitution and Transformed Bounds

Reverse the chain rule

The chain rule says that differentiating $F(g(x))$ produces $F'(g(x))g'(x)$ . Substitution reverses this structure. Choose $u=g(x)$ , so $du=g'(x)dx$ , and rewrite every remaining part in terms of $u$ .

A good substitution simplifies the integrand and has a derivative that can be supplied by the remaining factors. Merely renaming a complicated expression is not enough if unrelated $x$ terms remain and cannot be eliminated.

Visual guide

VISUAL GUIDEChange the variable and its interval

Original variable x

New variable u = x²

The area under 2x on [0, 1] equals 1. Under u = x², du = 2x dx, the same integral becomes the area under 1 on u ∈ [0, 1]. Substitution changes both the integrand measure and the endpoint labels.

Worked example: a composition with its derivative

For $\int2x\cos(x^2)dx$ , set $u=x^2$ . Then $du=2x\,dx$ , and the integral becomes $\int\cos u\,du=\sin u+C$ . Substituting back gives $\sin(x^2)+C$ . Differentiation confirms both the outer function and the required $2x$ factor.

For $\int x\cos(x^2)dx$ , the needed factor is twice what is available. Compensate with a constant: the result is $\tfrac12\sin(x^2)+C$ . Constant factors can be adjusted; variable factors cannot simply be invented.

Worked example: change the endpoints too

Evaluate $\int_0^1 2x/(1+x^2)\,dx$ . Choose $u=1+x^2$ and $du=2x\,dx$ . The old endpoint $x=0$ becomes $u=1$ , and $x=1$ becomes $u=2$ . Therefore

\int_0^1\frac{2x}{1+x^2}dx=\int_1^2\frac{du}{u}=\ln2.

An equally valid route is to find $\ln(1+x^2)$ and then use the original $x$ endpoints. Mixing a $u$ antiderivative with $x$ endpoints is invalid because they refer to different coordinates.

Reversing orientation

If $u$ decreases as $x$ increases, the transformed bounds reverse order and the differential carries the appropriate sign. For $\int_0^1(1-x)^2dx$ , let $u=1-x$ . Then $dx=-du$ and the bounds go from $1$ to $0$ , giving $-\int_1^0u^2du=1/3$ .

For the standard substitution theorem, a continuously differentiable inner function and continuous outer integrand on the relevant ranges are sufficient. More general changes of variable require care about domains and branches; trigonometric substitution later makes these choices explicit.

Practice

Integrate $3x^2e^{x^3}$ .
Evaluate $\int_0^1 x\sqrt{1+x^2}\,dx$ .
Why is $u=x^2$ not immediately enough to simplify $\int\cos(x^2)dx$ to $\int\cos u\,du$ ?

Show worked solutions

With $u=x^3$ , the answer is $e^{x^3}+C$ .
$u=1+x^2$ gives $\tfrac12\int_1^2u^{1/2}du=(2\sqrt2-1)/3$ .
The differential is $du=2x\,dx$ , and no $2x$ factor is present. Dropping it changes the integral; this example has no elementary antiderivative.