Every substitution is the chain rule read backwards — rename the messy inner function, and the integral simplifies itself.
10 min read5 worked examples
Formalising the reverse chain rule
The previous lesson spotted composite-function patterns by inspection, whenever the "extra" factor was exactly $f'(x)$. Integration by substitution makes that idea rigorous and mechanical — introducing a new variable $u$ for the inner function, and carefully converting every part of the integral, including $dx$ itself, into $u$. This handles cases the reverse chain rule can't reach directly, where the outer factor isn't a clean constant multiple of the inner function's derivative.
The substitution is chosen so that $u$ and its derivative $\dfrac{du}{dx}$ together account for everything inside the integral. Since $\dfrac{du}{dx}$ behaves algebraically like a fraction here, it can be rearranged to give $dx$ in terms of $du$, allowing the whole integral to be rewritten purely in terms of $u$.
The method
To evaluate $\displaystyle\int f(x)\,dx$ using the substitution $u=g(x)$:
1) Differentiate to get $\dfrac{du}{dx}$, then rearrange to write $dx = \dfrac{du}{\,du/dx\,}$. 2) Replace every $x$-term (including $dx$) with its $u$-equivalent. 3) Integrate with respect to $u$. 4) Substitute $u=g(x)$ back in, to give the answer in terms of $x$.
Definite integrals: changing the limits
For a definite integral $\int_a^b f(x)\,dx$, there are two valid approaches once a substitution $u=g(x)$ is introduced. Either integrate to get an antiderivative in terms of $u$, substitute back to $x$, and apply the original limits — or, more efficiently, convert the limits themselves into $u$-values using $u=g(x)$, and evaluate the entire definite integral in terms of $u$ without ever substituting back.
Changing the limits
If $u=g(x)$, then when $x=a$, $u=g(a)$; when $x=b$, $u=g(b)$. The new limits are $g(a)$ and $g(b)$ — and crucially, they don't need to be reconverted, since the final numerical answer is the same regardless of which variable it was computed in.
Worked example 1
Substitution with a linear inner function
Find $\displaystyle\int (3x+1)^5\,dx$.
MethodLet u equal the inner function, rewrite dx, then integrate
1
Let $u=3x+1$.
This is the expression being raised to a power — the natural inner function.
Differentiating $\dfrac{(3x+1)^6}{18}$ using the chain rule gives $\dfrac{6(3x+1)^5 \cdot 3}{18} = (3x+1)^5$ — confirming the answer, and showing exactly why substitution reverses the chain rule.
Worked example 2
Definite integral with changed limits
Find $\displaystyle\int_0^2 x(x^2+1)^3\,dx$ using the substitution $u=x^2+1$.
MethodConvert the limits into u-values, then integrate entirely in u
Rewrite entirely in u — the cos x · dx pair becomes du directly.
4
$=\dfrac{u^4}{4}+c$.
Integrate using the power rule.
5
$=\dfrac{\sin^4x}{4}+c$.
Substitute back to x.
▪ solved
Look deeper — spotting hidden substitutions
No algebraic composite function like $(3x+1)^5$ appears here — yet the same principle applies, since $\sin x$ (raised to a power) and its derivative $\cos x$ both sit inside the integrand. Substitution works whenever an integral contains a function and (a multiple of) its own derivative, whatever form that function takes.
Worked example 5
A trigonometric substitution, giving an inverse sine · Challenge
Find $\displaystyle\int \dfrac{1}{\sqrt{4-x^2}}\,dx$ using the substitution $x=2\sin\theta$.
MethodSubstitute x in terms of θ so the square root simplifies via an identity
1
Let $x=2\sin\theta$, so $\dfrac{dx}{d\theta}=2\cos\theta \Rightarrow dx = 2\cos\theta\,d\theta$.
This time it's x, not u, that's substituted — chosen to match the $\sqrt{4-x^2}$ shape.
2
$4-x^2 = 4-4\sin^2\theta = 4(1-\sin^2\theta) = 4\cos^2\theta$, so $\sqrt{4-x^2}=2\cos\theta$.
Apply $\sin^2\theta+\cos^2\theta=1$ to clear the square root entirely.
An integrand containing $\sqrt{a^2-x^2}$ → substitute $x=a\sin\theta$, giving $\displaystyle\int\frac{dx}{\sqrt{a^2-x^2}}=\arcsin\!\Big(\frac{x}{a}\Big)+c$.
An integrand containing $a^2+x^2$ → substitute $x=a\tan\theta$, giving $\displaystyle\int\frac{dx}{a^2+x^2}=\frac1a\arctan\!\Big(\frac{x}{a}\Big)+c$.
Common mistake
Forgetting to convert $dx$ into an expression involving $du$, and instead substituting only the "obvious" part of the integrand. Every $x$ — including the one hidden inside $dx$ — must be rewritten in terms of $u$ before integrating; skipping this step leaves a mixed-variable expression that cannot be integrated correctly.
Check yourself
Find $\displaystyle\int_0^1 \dfrac{6x^2}{x^3+1}\,dx$. (Hint: substitute u = x³ + 1 and change the limits.)
Method
1
Let $u=x^3+1$, so $\dfrac{du}{dx}=3x^2 \Rightarrow 6x^2\,dx = 2\,du$.