U-Substitution
This page follows the JoeCMath video Only 3 Examples to Finally Learn U-Substitution. The video uses three integrals to show how substitution reverses the chain rule: choose the inside expression, rewrite the matching derivative as $du$, integrate in terms of $u$, and then return to $x$.
Main Idea
U-substitution is useful when an integral has a function inside another function and also has the derivative of that inside function, possibly with a constant adjustment.
The video keeps that idea concrete. Each example asks for a good choice of $u$, rewrites the nearby differential piece as $du$, and then changes the integral into a simpler one.
Watch this section: introduction at 0:00.
Chain Rule Revisited
The video starts by connecting u-substitution back to the chain rule.
If a derivative has the form
\[\frac{d}{dx}\left[f(g(x))\right]=f'(g(x))g'(x),\]then u-substitution works in the reverse direction during integration. The inside function becomes $u$, and the derivative of that inside function becomes part of $du$.
For the chain-rule example shown in the video, if
\[f(x)=\cos(x^5),\]then
\[f'(x)=-\sin(x^5)\cdot 5x^4.\]That extra $5x^4$ factor is the clue u-substitution is built to use.
Watch this section: chain rule review at 0:24.
Example 1: The Inside Derivative Is Already There
The first worked example is
\[\int (x^5+1)^3 5x^4\,dx.\]The inside expression is
\[u=x^5+1.\]Differentiate both sides:
\[du=5x^4\,dx.\]That means the whole integral rewrites cleanly as
\[\int (x^5+1)^3 5x^4\,dx=\int u^3\,du.\]Now integrate in terms of $u$:
\[\int u^3\,du=\frac{u^4}{4}+C.\]Finally, replace $u$ with the expression it came from:
\[\frac{(x^5+1)^4}{4}+C.\]This is the cleanest version of the pattern because the $du$ piece is already sitting in the integral.
Watch this section: first example at 1:00.
U-Substitution Steps
The video pauses after the first example to make the process explicit.
- Select $u$ to make the integral simpler.
- Find $du$ and make sure the differential piece can be accounted for.
- Substitute, integrate in terms of $u$, and then replace $u$ with the original expression.
The important habit is not picking $u$ randomly. The best choice usually comes from the inside expression of a composite function.
Watch this section: u-substitution steps at 2:54.
Example 2: Adjusting a Constant Factor
The second worked example is
\[\int (3x-1)^6\,dx.\]Choose the inside expression:
\[u=3x-1.\]Then
\[du=3\,dx,\]so
\[dx=\frac{1}{3}\,du.\]The integral becomes
\[\int (3x-1)^6\,dx=\frac{1}{3}\int u^6\,du.\]Integrate:
\[\frac{1}{3}\int u^6\,du=\frac{1}{3}\cdot\frac{u^7}{7}+C=\frac{u^7}{21}+C.\]Return to $x$:
\[\frac{(3x-1)^7}{21}+C.\]This example shows that the derivative of the inside expression does not have to appear perfectly at first. A constant factor can be moved into the integral or pulled outside.
Watch this section: second example at 3:14.
Example 3: The Logarithm Is the Inside Function
The last worked example is
\[\int \frac{\sin(\ln(x))}{x}\,dx.\]The inside expression is
\[u=\ln(x).\]Then
\[du=\frac{1}{x}\,dx.\]So the integral becomes
\[\int \sin(u)\,du.\]Integrate:
\[\int \sin(u)\,du=-\cos(u)+C.\]Replace $u$ with $\ln(x)$:
\[-\cos(\ln(x))+C.\]The clue is the factor $\frac{1}{x}\,dx$. It matches the derivative of $\ln(x)$, so the logarithm can become the substitution variable.
Watch this section: third example at 5:02.
Timestamp Guide
Related Calculus Work
The derivative pattern behind this lesson is the chain rule. For that background, use the chain rule derivative examples.
You can also browse calculus derivative notes or return to the new u-substitution topic page.
The video description links to the JoeCMath integration playlist: watch the integration playlist on YouTube.