Recall the chain rule:
[latex]\boxed{\cfrac{d}{dx} f(g) = f'(g) \cdot g'}[/latex]
The method of “[latex]u[/latex]-substitution” is a way of doing integral problems that undo the chain rule. It also helps deal with constants that crop up.
- Identify an “inside” function whose derivative is multiplied on the outside, possibly with a different constant. Call this “inside” function [latex]u[/latex].
- Compute [latex]\frac{du}{dx}[/latex] and solve for [latex]dx[/latex].
- Use substitution to replace [latex]x \to u[/latex] and [latex]dx \to du[/latex], and cancel any remaining [latex]x[/latex] terms if possible.
- Integrate with respect to [latex]u[/latex]. If at this point you still have any [latex]x[/latex]s in your problem, either you made a mistake or the method of [latex]u[/latex]-substitution will not work for this problem.
- Substitute back the [latex]x[/latex]s back into the answer before evaluating the definite integral.
Let’s do some examples.
We will follow the steps of [latex]u[/latex] substitution.
- In this case, the “inside function” is [latex]u = -5x[/latex].
- If we compute [latex]\frac{du}{dx}[/latex], we see the derivative of [latex]-5x[/latex] is [latex]-5[/latex]. Hence [latex]\frac{du}{dx} = -5,[/latex] and we have solving for [latex]dx[/latex]
\begin{align*}
\frac{du}{dx} & = -5 \\
du & = -5 dx \\
-\frac{1}{5} du & = dx \end{align*} - Going back to the original problem and using substitution [latex]u = -5x[/latex] and [latex]dx = -\frac{1}{5} du[/latex], we thus have:
\begin{align*}
\int_0^3 e^{-5x}dx & \int_0^3 e^u dx \\
& = \int_0^3 e^u \left(- \frac{1}{5} du\right) \\
& = \int_0^3 -\frac{1}{5} e^udu.
\end{align*} - Integrating with respect to [latex]u[/latex],
\begin{align*}
\int_0^3 – \frac{1}{5} e^udu & = -\frac{1}{5} \int_0^3 e^udu \\
& = -\frac{1}{5} e^u \Big|_0^3
\end{align*} - We now substitute [latex]u = -5x[/latex]. We have
\begin{align*}
-\frac{1}{5} e^u \Big|_0^3 & = -\frac{1}{5}e^{-5x} \Big|_0^3 \\
& = \left(-\frac{1}{5}e^{-15}\right) – \left(-\frac{1}{5}e^{-5(0)} \right) \\
& = -\frac{e^{-15}}{5} + \frac{1}{5} \\
& = \frac{1 – e^{-15}}{5}\\
& \approx \boxed{0.2}
\end{align*}
There we go.
Again, we will go through the steps of [latex]u[/latex]-substitution.
- The inside function in this case is [latex]x^2 + 1[/latex]. We can see that the derivative is [latex]2x[/latex], and this is good since there is an [latex]x[/latex] multiplied out in front (the [latex]2[/latex] is just a constant we can deal with.) Set [latex]u = x^2 + 1[/latex].
- We see [latex]\frac{du}{dx} = 2x[/latex], and hence solving for [latex]dx[/latex] we have [latex]\frac{du}{2x} = dx[/latex].
- Subbing in [latex]u = x^2 + 1[/latex] and [latex]dx = \frac{du}{2x}[/latex], we have
\begin{align*}
\int x (x^2 + 1)^7dx & = \int x u^7dx \\
& = \int x u^7 \left( \frac{du}{2x} \right) \\
& = \int \frac{x}{2x} u^7du \\
& = \int \frac{1}{2} u^7du
\end{align*}
Great, the [latex]x[/latex]s are all gone! - We can now integrate with respect to [latex]u[/latex].
\begin{align*}
\int \frac{1}{2} u^7du & = \frac{1}{2} \int u^7du \\
& = \frac{1}{2} \frac{u^8}{8} \\
& = \frac{u^8}{16}
\end{align*} - Finally, we sub in the [latex]x[/latex]s again using [latex]u = x^2 + 1[/latex].
\begin{align*}
\frac{u^8}{16} & = \boxed{\frac{(x^2 + 1)^8}{16} + C}
\end{align*}
Since this is an indefinite integral, we add the constant of integration.
Again, we will go through the steps of [latex]u[/latex]-substitution.
- The inside function in this case is [latex]\ln(x)[/latex]. We can see that the derivative is [latex]\frac{1}{x}[/latex], and this is good since there is an [latex]x[/latex] dividing the rest of the problem. Set [latex]u = \ln(x)[/latex].
- We see [latex]\frac{du}{dx} =\frac{1}{x}[/latex], and hence solving for [latex]dx[/latex] we have [latex]\frac{du}{\frac{1}{x}} = dx[/latex].
- Subbing in [latex]u =\ln(x)[/latex] and [latex]dx = \frac{du}{\frac{1}{x}}[/latex], we have
\begin{align*}
\int \frac{8 (\ln(x))^3}{x}dx & = \int \frac{8 u^3}{x} \cdot \frac{du}{\frac{1}{x}} \\
& = \int \frac{8 u^3}{x \cdot \frac{1}{x}}du \\
& = \int \frac{8 u^3}{1}du \\
& = \int 8 u^3du \\
\end{align*}
Great, the [latex]x[/latex]s are all gone! - We can now integrate with respect to [latex]u[/latex].
\begin{align*}
\int 8 u^3du & = 8 \cdot \frac{1}{4} u^4 \\
& = 2 u^4
\end{align*} - Finally, we sub in the [latex]x[/latex]s again using [latex]u = \ln(x)[/latex].
\begin{align*}
\int \frac{8(\ln(x))^3}{x}dx & = \boxed{2 (\ln(x))^4 + C}.
\end{align*}
If the inside function is linear, the [latex]u[/latex]-substitution is much simpler, and there is even a formula for it (just like in the [latex]\int e^{-5x}dx[/latex] example above). By the chain rule with [latex]g = mx + b[/latex] and [latex]g' = m[/latex], we have
\begin{align*}
\frac{d}{dx} \frac{1}{m} f(mx + b) & = \frac{1}{m} \frac{d}{dx} f(mx + b) \\
& = \frac{1}{m} \left( f'(g) \cdot g’ \right) \\
& = \frac{1}{m} f'(mx + b) \cdot m \\
& = \frac{m}{m} f'(mx + b) \\
& = f'(mx + b)
\end{align*}
If we integrate both sides of this equation, we have the following useful rule which I call the “chain rule shortcut”:
[latex]\boxed{\int f'(mx + b)dx = \frac{1}{m} f(mx + b)}[/latex]
This is especially important if [latex]f(x) = e^{mx}[/latex]. In this case, we have
[latex]\boxed{\int e^{mx}dx = \frac{1}{m} e^{mx}}[/latex]
In other words, you integrate just like normal without any [latex]u[/latex]-substitution, and then add a [latex]\frac{1}{m}[/latex] factor for the fact that you have [latex]mx + b[/latex] inside the function instead of just an [latex]x[/latex].
Let’s see a couple of examples of this:
-
Find [latex]\int (6x - 3)^4dx[/latex].
Using the inverse power rule, we see this becomes [latex]\frac{(6x - 3)^5}{5}[/latex]. However, we need that [latex]\frac{1}{m}[/latex] factor since the problem has a [latex]6x - 3[/latex] in it. So the final answer is
[latex]\frac{1}{6} \frac{(6x - 3)^5}{5} = \frac{(6x - 3)^5}{30} + C.[/latex]
You’d get the same thing doing a full [latex]u[/latex]-substitution with [latex]u = 6x - 3[/latex]. This way, though, you save some time by just multiplying by [latex]\frac{1}{6}[/latex].
-
Find [latex]\int_{-2}^{-1} (3x + 5)^7dx[/latex].
Again, we use the [latex]u[/latex]-sub shortcut — we just need to do power rule and remember a [latex]\frac{1}{m}[/latex] factor, which in this case is [latex]\frac{1}{3}[/latex].
\begin{align*}
\int_{-2}^{-1} (3x + 5)^7dx & = \frac{1}{3} \frac{(3x + 5)^8}{8} \Big|_{-2}^{-1} \\
& = \frac{(3x + 5)^8}{24} \Big|_{-2}^{-1} \\
& = \left( \frac{(3(-1) + 5)^8}{24} \right) – \left( \frac{(3(-2) + 5)^8}{24} \right) \\
& = \frac{256}{24} – \frac{1}{24} \\
& = \frac{255}{24} \\
& = \frac{83}{8}
\end{align*} -
Find [latex]\int e^{0.05t}dt[/latex].
The antiderivative here is just [latex]\frac{1}{0.05} e^{0.05t}[/latex]. We can plug [latex]1/0.05[/latex] into a calculator and get [latex]20[/latex], so the answer is [latex]\boxed{20 e^{0.05t} + C}[/latex].
