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