U-Substitution With a Constant Inside
This page follows the JoeCMath video The Simplest U-Substitution You’ll Ever Learn. The video uses one compact example to show why a constant inside the function changes the antiderivative by a reciprocal factor.
Main Idea
The video starts with
\[\int \cos(3x)\,dx.\]The outside function is cosine, and the inside function is $3x$. Since $3x$ is just a constant times the variable, the example becomes a clean place to see how u-substitution reverses the chain rule.
Watch this section: integrating $\cos(3x)$ at 0:00.
The First Guess
A natural first guess is
\[\sin(3x)+C,\]because the antiderivative of cosine is sine. The video immediately checks that guess by differentiating it:
\[\frac{d}{dx}\left[\sin(3x)\right]=3\cos(3x).\]That derivative is too large. It produces $3\cos(3x)$ instead of the original $\cos(3x)$.
Watch this section: checking the first guess at 0:29.
Fixing The Factor Of 3
The first guess only misses by a constant factor. To cancel the extra $3$ created by the chain rule, the video multiplies the guess by $\frac{1}{3}$:
\[\frac{d}{dx}\left[\frac{1}{3}\sin(3x)\right] =\frac{1}{3}\cdot 3\cos(3x) =\cos(3x).\]So the antiderivative is
\[\int \cos(3x)\,dx=\frac{1}{3}\sin(3x)+C.\]The missing adjustment is not a new trick; it is the chain-rule factor being undone.
Watch this section: fixing the answer with $\frac{1}{3}$ at 1:05.
Doing It With U-Substitution
The video then solves the same integral using substitution. Start with the inside function:
\[u=3x.\]Differentiate and solve for $dx$:
\[\frac{du}{dx}=3, \qquad du=3\,dx, \qquad dx=\frac{1}{3}\,du.\]Now substitute $u$ for $3x$ and $\frac{1}{3}\,du$ for $dx$:
\[\int \cos(3x)\,dx =\int \cos(u)\cdot \frac{1}{3}\,du =\frac{1}{3}\int \cos(u)\,du.\]Integrate in terms of $u$ and then replace $u$ with $3x$:
\[\frac{1}{3}\sin(u)+C =\frac{1}{3}\sin(3x)+C.\]This is where the $\frac{1}{3}$ comes from inside the formal u-substitution work.
Watch this section: solving again with u-substitution at 1:45.
The Shortcut For $f(kx)$
After the full substitution, the video names the pattern. When the inside is $kx$ with $k\ne 0$, and there is no extra $x$ factor outside, take the outside antiderivative, keep the inside expression, and divide by $k$.
One way to write that is:
\[\int f(kx)\,dx=\frac{1}{k}F(kx)+C, \qquad F'(x)=f(x),\ k\ne 0.\]This shortcut is the reverse-chain-rule idea from the example. Differentiating $F(kx)$ would multiply by $k$, so integrating divides by $k$.
Watch this section: shortcut for $f(kx)$ at 3:02.
Two Quick Examples
The video closes by applying the same idea to two fast examples.
For $e^{5x}$, keep $5x$ in the exponent and divide by $5$:
\[\int e^{5x}\,dx=\frac{1}{5}e^{5x}+C.\]For $\sin(4x)$, the outside antiderivative is $-\cos(x)$, so keep $4x$ inside and divide by $4$:
\[\int \sin(4x)\,dx=-\frac{1}{4}\cos(4x)+C.\]Both examples are doing the same thing as $\int \cos(3x)\,dx$: they undo the constant factor that the chain rule would create during differentiation.
Watch these sections: $e^{5x}$ at 3:20 and $\sin(4x)$ at 3:35.
Timestamp Guide
Related Calculus Work
For a longer u-substitution walkthrough with three examples, use the u-substitution companion.
If the outside function is a power of $x$ instead, review the integration power rule. For why indefinite integral answers include $+C$, use the constant of integration notes.
The video description also links to the full JoeCMath u-substitution playlist: watch the u-substitution playlist on YouTube.