A-Level Maths Integration by Substitution Step by Step
Integration by substitution reverses the chain rule. It is particularly useful when an integral contains a complicated expression and another factor that resembles its derivative. Replacing the inner expression with a new variable often turns a difficult-looking integral into a familiar power or logarithm rule. The most important skill is transforming the whole integrand, including the differential.
Spot an inner function and its derivative
Consider ∫2x(x² + 1)³ dx. The inner expression x² + 1 has derivative 2x, which appears elsewhere in the integrand. Set u = x² + 1. Then du/dx = 2x, so du = 2x dx. The integral becomes ∫u³ du. This transformation works because every appearance of x relevant to the integrand is replaced, not merely the expression inside the brackets.
Integrate and substitute back
Integrating u³ gives u⁴/4 + C. Since u = x² + 1, the answer is (x² + 1)⁴/4 + C. Differentiate this answer to check: the power rule gives (x² + 1)³ multiplied by the derivative 2x, recovering the original integrand. The constant C is required for an indefinite integral because many antiderivatives differ by an additive constant.
When the inner derivative is not an exact factor
For ∫cos(3x) dx, let u = 3x, so du = 3 dx and dx = du/3. The integral becomes (1/3)∫cos u du = (1/3)sin u + C. Substituting back gives (1/3)sin(3x) + C. A common error is to forget the factor one third. Differentiating sin(3x) would produce 3cos(3x), showing exactly why the adjustment is necessary.
Substitution leading to a logarithm
For ∫x/(x² + 4) dx, set u = x² + 4 and du = 2x dx. Therefore x dx = du/2. The integral becomes (1/2)∫1/u du = (1/2)ln|u| + C, giving (1/2)ln(x² + 4) + C because x² + 4 is positive. The method reveals a logarithmic structure that may not be obvious in the original expression.
Change the limits for a definite integral
Evaluate ∫ from x = 0 to x = 1 of 2x(x² + 1)² dx. With u = x² + 1, the lower limit becomes u = 1 and the upper becomes u = 2. The new integral is ∫ from 1 to 2 of u² du = [u³/3] from 1 to 2 = (8 − 1)/3 = 7/3. Because the limits have changed to u-values, there is no need to substitute x back before evaluating.
Common substitution errors
Do not change the integrand without changing dx. Do not mix x-based limits with an integrand expressed in u. If you keep the original limits, integrate and substitute x back before evaluating; if you transform the limits, remain in u throughout. Make a derivative check whenever possible. A substitution should simplify the whole integral; if x terms remain awkwardly mixed with u, reconsider the choice or algebra.
Frequently asked questions
Does integration by substitution always work?
It is useful when the integrand has a suitable composite-function structure, but not every integral simplifies through a straightforward substitution.
Should I change the limits of definite integrals?
You can change the limits to the new variable, or substitute back before evaluating, but do not mix the two methods.
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