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.

\(u\)-substitution:

1. Identify an “inside” function whose derivative is multiplied on the outside, possibly with a different constant. Call this “inside” function \(u\).
2. Compute \(\frac{du}{dx}\) and solve for \(dx\).
3. Use substitution to replace \(x \to u\) and \(dx \to du\), and cancel any remaining \(x\) terms if possible.
4. 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.
5. Substitute back the \(x\)s back into the answer before evaluating the definite integral.

Let’s do some examples.

\(u\)-substitution

Find \(\int_0^3 e^{(-5x)}dx\).

We will follow the steps of \(u\) substitution.

1. In this case, the “inside function” is \(u = -5x\).
2. If we compute \(\frac{du}{dx}\), we see the derivative of \(-5x\) is \(-5\). Hence \(\frac{du}{dx} = -5,\) and we have solving for \(dx\)

   \[\begin{align*} \frac{du}{dx} & = -5 \\ du & = -5 dx \\ -\frac{1}{5} du & = dx \end{align*}\]

3. Going back to the original problem and using substitution \(u = -5x\) and \(dx = -\frac{1}{5} du\), we thus have:

   \[\begin{align*} \int_0^3 e^{-5x}dx & \int_0^3 e^u dx \\ & = \int_0^3 e^u \left(- \frac{1}{5} du\right) \\ & = \int_0^3 -\frac{1}{5} e^udu. \end{align*}\]

4. Integrating with respect to \(u\),

   \[\begin{align*} \int_0^3 - \frac{1}{5} e^udu & = -\frac{1}{5} \int_0^3 e^udu \\ & = -\frac{1}{5} e^u \Big|_0^3 \end{align*}\]

5. 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.

\(u\)-substitution

Find the indefinite integral \(\int x(x^2 + 1)^7dx\).

Again, we will go through the steps of \(u\)-substitution.

1. 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 \(u = x^2 + 1\).
2. We see \(\frac{du}{dx} = 2x\), and hence solving for \(dx\) we have \(\frac{du}{2x} = dx\).
3. 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!

4. We can now integrate with respect to \(u\).

   \[\begin{align*} \int \frac{1}{2} u^7du & = \frac{1}{2} \int u^7du \\ & = \frac{1}{2} \frac{u^8}{8} \\ & = \frac{u^8}{16} \end{align*}\]

5. 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.

\(u\)-substitution

Find the indefinite integral \(\int \frac{8(\ln(x))^3}{x}dx\).

Again, we will go through the steps of \(u\)-substitution.

1. 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 \(u = \ln(x)\).
2. We see \(\frac{du}{dx} =\frac{1}{x}\), and hence solving for \(dx\) we have \(\frac{du}{\frac{1}{x}} = dx\).
3. 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!

4. We can now integrate with respect to \(u\).

   \[\begin{align*} \int 8 u^3du & = 8 \cdot \frac{1}{4} u^4 \\ & = 2 u^4 \end{align*}\]

5. 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:

Chain rule shortcut

• 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 \(\frac{1}{6}\).

• 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 \(\frac{1}{3}\).

  \[\begin{align*} \int_{-2}^{-1} (3x + 5)^7dx & = \frac{1}{3} \frac{(3x + 5)^8}{8} \Big|_{-2}^{-1} \\ & = \frac{(3x + 5)^8}{24} \Big|_{-2}^{-1} \\ & = \left( \frac{(3(-1) + 5)^8}{24} \right) - \left( \frac{(3(-2) + 5)^8}{24} \right) \\ & = \frac{256}{24} - \frac{1}{24} \\ & = \frac{255}{24} \\ & = \frac{83}{8} \end{align*}\]

• 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 \(\boxed{20 e^{0.05t} + C}\).