U-Substitution With a Constant Inside

July 12, 2026

Simple overview of finding integrals of where we have a composition of functions and the inner function is linear relationship between a constant and our variable of integration.

Main Idea

Consider

\[\int \cos(3x)dx\]

The outside function is cosine, and the inside function is $3x$.

Let’s consider what the antiderivative is in this scenario.

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.

Look what happens when we differentiate $sin(3x)$

\[\frac{d}{dx}\left[\sin(3x)\right]=3\cos(3x)\]

There is a $3$ in front of $cos(3x)$ and since $\frac{d}{dx}\left[\sin(3x)\right] = 3cos(3x)\neq cos(3x)$ we know $\sin(3x)+C$ is not our antiderivative.

Watch this section: checking the first guess at 0:29.

Fixing The Factor Of 3

To cancel the extra $3$ created by the chain rule, we can multiply the last guess by $\frac{1}{3}$ and see if that works

\[\frac{d}{dx}\left[\frac{1}{3}\sin(3x)\right]=\frac{1}{3}\cdot 3\cos(3x)=\cos(3x)\]

So the antiderivative is

\[\int \cos(3x) dx=\frac{1}{3}\sin(3x)+C\]

Watch this section: fixing the answer with $\frac{1}{3}$ at 1:05.

Doing It With U-Substitution

Let’s now find it using u-substitution

Let

\[u=3x\]

Differentiate and solve for $dx$

\[\frac{du}{dx}=3 \qquad du=3dx \qquad dx=\frac{1}{3}du\]

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 $3x$:

\[\frac{1}{3}\sin(u)+C = \frac{1}{3}\sin(3x)+C\]

Notice we get the same result using u-substitution.

Watch this section: solving again with u-substitution at 1:45.

The Shortcut For $f(kx)$

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 $k$.

Consider

\[\int f(kx)dx=\frac{1}{k}F(kx)+C \qquad F'(x)=f(x)\text{ where } k\ne 0\]

Watch this section: shortcut for $f(kx)$ at 3:02.

Timestamp Guide

Section	Main idea	Video
$\int \cos(3x)\,dx$	Identify the outside cosine function and the inside $3x$.	0:00
First guess	Try $\sin(3x)+C$ before checking it.	0:19
Derivative check	Differentiating $\sin(3x)$ creates an extra factor of $3$.	0:29
Fixing the factor	Multiply by $\frac{1}{3}$ so the derivative returns $\cos(3x)$.	1:05
U-substitution setup	Let $u=3x$, then rewrite $dx$ as $\frac{1}{3}\,du$.	1:45
Substitute and integrate	Change the integral to $\frac{1}{3}\int \cos(u)\,du$.	2:32
Shortcut	For $f(kx)$, keep the inside and divide by $k$.	3:02
Example: $e^{5x}$	Divide the antiderivative by $5$.	3:20
Example: $\sin(4x)$	Use $-\cos(4x)$ and divide by $4$.	3:35
Reverse chain rule	The constant division undoes the chain-rule multiplier.	3:53

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.