U-Substitution and Changing the Bounds

You’ll be able to

Choose a substitution by looking for an inner function whose derivative is present as a factor
Convert the limits of integration when substituting in a definite integral
Recognize when substitution alone will not work and another technique is needed

Substitution undoes the chain rule

Every chain-rule derivative has the shape f′(g(x))·g′(x), so every integral of that shape can be reversed. The move is to set u = g(x), the inner function, and du = g′(x) dx. The substitution succeeds exactly when the leftover factor in the integrand is du up to a constant — that is the test to apply before committing. For ∫ 2x·cos(x²) dx, u = x² gives du = 2x dx, and the integral becomes ∫ cos u du = sin u + C = sin(x²) + C.

Change the bounds, or change back

For a definite integral there are two legitimate finishes and one common error. Either convert the limits into u-values and evaluate entirely in u — ∫₀² 2x·cos(x²) dx becomes ∫₀⁴ cos u du — or antidifferentiate, substitute back to x, and use the original limits. What you may never do is keep the x-limits while evaluating in u. Converting the limits is the safer habit because it removes the back-substitution step entirely.

Choosing a technique

BC gives you four integration tools, and knowing which to reach for is itself examined. Substitution when an inner function's derivative appears as a factor. Integration by parts when the integrand is a product of unlike types — a polynomial times an exponential or a logarithm. Partial fractions when the integrand is a rational function whose denominator factors. And plain rewriting — expanding, splitting a fraction, using a trig identity — surprisingly often. If a substitution leaves a stray x that will not convert, that is the signal to switch tools rather than to force it.

Definite integral by substitution

∫ₐᵇ f(g(x))·g′(x) dx = ∫_(g(a))^(g(b)) f(u) du

The new limits are the old ones passed through g. Once the limits are converted, never substitute back.

Worked example

Evaluate ∫₁³ x/(x² + 1)² dx.

1.The inner function is x² + 1; its derivative 2x is present as a factor up to the constant 2, so substitution will work.
2.Let u = x² + 1, so du = 2x dx, which means x dx = du/2.
3.Convert the limits: when x = 1, u = 2; when x = 3, u = 10.
4.The integral becomes (1/2)∫₂¹⁰ u⁻² du = (1/2)[−u⁻¹] from 2 to 10.
5.Evaluate: (1/2)[(−1/10) − (−1/2)] = (1/2)(2/5) = 1/5.

Answer: 1/5. Converting the limits to u = 2 and u = 10 avoids any back-substitution; using x = 1 and x = 3 on the u-antiderivative would give a wrong answer.

Watch out

The most expensive substitution error is silent: finishing in u and then plugging in the original x-limits. Write the new limits on the integral sign the moment you choose u, so there is nothing left to forget.

Checkpoint

For ∫₀^(π/2) sin⁴(x)cos(x) dx, which substitution and limits are correct?

Checkpoint

Which integral is best handled by integration by parts rather than substitution?

On the exam

BC Unit 6 is the largest unit on the exam, and much of its weight is technique selection rather than execution. When practicing, say which tool you are choosing and why before doing any algebra.

Answer the 2 checkpoints as you read.