DE-BOOK Integration by parts

Section 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{50} \end{equation}

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Let’s consider the following example.

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🌌 Example 325. 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*}

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Now you try some.

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Evaluate each of the following integrals. Use proper notation.

\(\ds \int (x - 1)e^x \ dx \qquad\)
Solution.

We choose \(u\) and \(dv\) as follows:

\begin{equation*} u = x-1 \hspace{2cm} dv = e^x dx. \end{equation*}

Then we find \(du\) by taking the derivative of \(u\) and \(v\) by taking the antiderivative of \(dv\text{:}\)

\begin{equation*} du = dx \hspace{2cm} v = e^x. \end{equation*}

Thus we have:

\begin{align*} \int (x - 1)e^x dx \amp = \int \ub{(x - 1)}_{u} \ub{e^x dx}_{dv} \\ \amp = \int u\cdot dv \\ \amp = u\cdot v - \int v \cdot du \\ \amp = (x-1)e^x - \int e^x dx \\ \amp = (x-1)e^x - e^x + C \\ \amp = xe^x - e^x - e^x + C \\ \amp = (x-2)e^x + C \end{align*}

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Answer.

\begin{equation*} (x-2)e^x + C \end{equation*}

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\(\ds \int x^2 \sin x \ dx \qquad\)
Solution.

We choose \(u\) and \(dv\) as follows:

\begin{equation*} u = x^2 \hspace{2cm} dv = \sin x \ dx. \end{equation*}

Then we find \(du\) by taking the derivative of \(u\) and \(v\) by taking the antiderivative of \(dv\text{:}\)

\begin{equation*} du = 2xdx \hspace{2cm} v = -\cos x. \end{equation*}

Thus we have:

\begin{align*} \int x^2 \sin x dx\amp = \int \ub{x^2}_{u} \ub{\sin x dx}_{dv} \\ \amp = \int u\cdot dv \\ \amp = u\cdot v - \int v \cdot du \\ \amp = x^2(-\cos x) - \int -\cos x \cdot 2x dx \\ \amp = -x^2\cos x + \int 2x\cos x dx \end{align*}

The remaining integral, \(\int 2x\cos x dx,\) is simpler than the one we started with, but we will need to do another integration by parts in order to evaluate it. Here we choose

\begin{equation*} u = 2x \hspace{2cm} dv = \cos x dx. \end{equation*}

Then we find \(du\) by taking the derivative of \(u\) and \(v\) by taking the antiderivative of \(dv\text{:}\)

\begin{equation*} du = 2dx \hspace{2cm} v = \sin x. \end{equation*}

Now we pick up where we left off:

\begin{align*} \int x^2 \sin x dx\amp = -x^2\cos x + \int 2x\cos x dx\\ \amp = -x^2\cos x + \int \ub{2x}_{u} \ub{\cos x dx}_{dv} \\ \amp = -x^2\cos x + \int u\cdot dv\\ \amp = -x^2\cos x + u\cdot v - \int v \cdot du \\ \amp = -x^2\cos x + 2x(\sin x) - \int \sin x \cdot 2 dx \\ \amp = -x^2\cos x + 2x\sin x - 2\int \sin x dx \\ \amp = -x^2\cos x + 2x\sin x - 2(-\cos x) +C \\ \amp = -x^2\cos x + 2x\sin x + 2\cos x +C \\ \amp = (2-x^2)\cos x + 2x\sin x +C \end{align*}

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Answer.

\begin{equation*} (2-x^2)\cos x + 2x\sin x +C \end{equation*}

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Integration by parts is a technique for transforming integrals in a way that makes them easier to solve. A good way to think about it is like redistributing "work" between two functions. One function takes on the derivative, while the other gets simplified through integration.

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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.

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Subsubsection Breaking Down the Integration by Parts Formula

Let’s break down the formula for integration by parts:

\begin{equation*} \int u\, dv = uv - \int v\, du \text{.} \end{equation*}

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Here’s how it works:

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\(u\) is a function that you choose to differentiate (it should get simpler when differentiated).

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\(dv\) is a part of the integrand that you choose to integrate (it should get easier when integrated).

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\(uv\) is the new term after applying the product of \(u\) and the integral of \(dv\text{.}\)

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\(\int v\, du\) is the remaining integral, now simpler than the original.

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Subsubsection Example: Applying Integration by Parts

Consider the integral:

\begin{equation*} \int t \, e^t \, dt \text{.} \end{equation*}

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We’ll apply integration by parts, following these steps:

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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 \(v = e^t\text{.}\)
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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*}

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Step 3: 🔗: Step 3: Solve the remaining integral:

\begin{equation*} \int e^t \, dt = e^t \text{.} \end{equation*}

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Step 4: 🔗: Step 4: Combine the results:

\begin{equation*} t \, e^t - e^t + C \text{.} \end{equation*}

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And that’s the final result:

\begin{equation*} \int t \, e^t \, dt = t \, e^t - e^t + C \text{.} \end{equation*}

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Subsubsection Laplace Transform and Integration by Parts: An Analogy

When applying the Laplace Transform, think of it as a way of "unwrapping" the derivatives of a function. 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.

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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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