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Section B.6 Integration by parts
Integration by parts may be a good choice when the integrand contains a product. Recall the formula for integration by parts.
\begin{equation}
\int u \cdot dv = u\cdot v - \int v \cdot du\tag{B.2}
\end{equation}
Let’s consider the following example.
🌌 Example 377 Evaluate \(\ds \int t^3 \ln t \ dt \) .
We choose \(u\) and \(dv\) as follows:
\begin{equation*}
u = \ln t \hspace{2cm} dv = t^3 \ dt.
\end{equation*}
Then we find \(du\) by taking the derivative of \(u\) and \(v\) by taking the antiderivative of \(dv\text{:}\)
\begin{equation*}
du = \frac{1}{t}dt \hspace{2cm} v = \frac{1}{4}t^4.
\end{equation*}
Thus we have:
\begin{align*}
\int t^3 \ln t dt \amp = \int \ln t \cdot t^3dt \\
\amp = \int \ub{\ln t}_{u} \cdot \ub{t^3 dt}_{dv} \\
\amp = \int u\cdot dv \\
\amp = u\cdot v - \int v \cdot du \\
\amp = \ln t \cdot \frac{1}{4}t^4 - \int \frac{1}{4}t^4
\cdot \frac{1}{t}dt \\
\amp = \frac{1}{4}t^4\ln t - \frac{1}{4}\int t^3 dt \\
\amp = \frac{1}{4}t^4\ln t - \frac{1}{4}\cdot \frac{1}{4}t^4 + C \\
\amp = \frac{1}{4}t^4\ln t - \frac{1}{16}t^4 + C
\end{align*}
Evaluate each of the following integrals. Use proper notation.
Checkpoint 378 .
\(\ds \int (x - 1)e^x \ dx \qquad\)
Answer .
\begin{equation*}
(x-2)e^x + C
\end{equation*}
Checkpoint 379 .
\(\ds \int x^2 \sin x \ dx \qquad\)
Answer .
\begin{equation*}
(2-x^2)\cos x + 2x\sin x +C
\end{equation*}
Integration by parts is a technique for transforming integrals to make them easier to solve. A good way to think about it is to think of redistributing "work" between two functions. One function takes on the derivative, while the other gets simplified through integration.
Imagine you have two workers. One is skilled at hard, repetitive tasks (like differentiating), and the other is better at creative problem-solving (like integrating). Integration by parts is like swapping their roles so that the right person handles the right task.
Example: Applying Integration by Parts.
Consider the integral:
\begin{equation*}
\int t \, e^t \, dt \text{.}
\end{equation*}
We’ll apply integration by parts, following these steps:
Step 1:
Identify
\(u\) and
\(dv\text{.}\) In this case, we choose
\(u = t\) and
\(dv = e^t \, dt\text{.}\) This makes
\(du = dt\) and
\(v = e^t\text{.}\)
Step 2:
Step 2: Substitute into the integration by parts formula:
\begin{equation*}
\int t \, e^t \, dt = t \, e^t - \int e^t \, dt \text{.}
\end{equation*}
Step 3:
Step 3: Solve the remaining integral:
\begin{equation*}
\int e^t \, dt = e^t \text{.}
\end{equation*}
Step 4:
Step 4: Combine the results:
\begin{equation*}
t \, e^t - e^t + C \text{.}
\end{equation*}
And that’s the final result:
\begin{equation*}
\int t \, e^t \, dt = t \, e^t - e^t + C \text{.}
\end{equation*}
Laplace Transform and Integration by Parts: An Analogy. When applying the Laplace Transform, think of it as a way to "unwrap" a function’s derivatives. Just like how you can redistribute "work" between functions using integration by parts, the Laplace Transform temporarily converts a differential equation into an algebraic one, allowing you to solve it more easily.
Once the problem is solved in the transformed space, we can "repackage" the function by applying the inverse Laplace Transform, revealing the solution in its original form.
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