Skip to main content

Section Integration by Parts

Integration by parts is an application of the product rule — essentially the product rule in reverse! It is typically used when the integrand is the product of two functions, one which you can easily anti-differentiate and one which becomes simpler (or at least no more complicated) when you differentiate.

Problem 5.11.

Find two functions \(f\) and \(g\) so that \(\dsp \int f(x) \cdot g(x) \; dx \neq \int f(x) \; dx \cdot \int g(x) \; dx\text{.}\)

Example 5.12.

Evaluate the integral \(\dsp \int x \ln (x) \; dx\text{.}\)

We begin by guessing a function which has \(x \ln(x)\) as a part of its derivative when we use the product rule. From the product rule for derivatives we have,

\begin{equation*} (x^{2} \ln x)' = 2x \ln x + x^{2} (\frac{1}{x}). \end{equation*}

Integrating both sides yields,

\begin{equation*} \dsp \int (x^{2} \ln x)' \; dx = \int 2 x \ln x \; dx + \int x \; dx. \end{equation*}

Now we have,

\begin{equation*} \dsp x^2 \ln x + k_1= 2 \int x \ln x \; dx + \frac{1}{2} x^2 + k_2 \end{equation*}

Where \(k_1\) and \(k_2\) are constants. Solving for the integral we wanted yields,

\begin{equation*} \dsp \int x \ln x \; dx = \frac{1}{2} x^2 \ln x - \frac{1}{4} x^2 + \frac{1}{2}(k_1-k_2). \end{equation*}
Problem 5.13.

Use the idea from the example to compute the following anti-derivatives.

  1. \(\dsp \int x^{45} \ln x \; dx\)

  2. \(\dsp \int x e^x \; dx\)

  3. \(\dsp \int x \cos(x) \; dx\)

Problem 5.14.

Let \(f\) and \(g\) be differentiable functions. Starting with the product rule for derivatives, show that

\begin{equation*} \int f(x)g'(x) \; dx = f(x) \cdot g(x) - \int g(x) f'(x) \; dx. \end{equation*}
Problem 5.15.

Evaluate each indefinite integral.

  1. \(\dsp \int x^{2} e^x \; dx\)

  2. \(\dsp \int x^{2} \sin(x) \; dx\)

  3. \(\dsp \int e^x \cos x \; dx\)

  4. \(\dsp \int (\ln x)^2 \; dx\)

  5. \(\dsp \int \sec^3 x \; dx\)

Problem 5.16.

Compute the derivatives of each.

  1. \(s(x) = \invsin(x)\) by differentiating the equation \(\sin(s(x))=x\) and solving for \(s'(x)\)

  2. \(c(x) = \invsec(x)\)

  3. \(t(x) = \invtan(x)\)

Now we address the integrals of the inverse trigonometric functions. Use integration by parts since you know the derivatives of each function.

Problem 5.17.

Compute the following indefinite integrals via integration by parts.

  1. \(\dsp \int \invsin(x) \; dx\) using \(\invsin(x) = \invsin(x) \cdot 1.\)

  2. \(\dsp \int \invtan(x) \; dx\)

  3. \(\dsp \int \invsec(x) \; dx\)

Writing \(f(x)\) as \(u\) and \(g(x)\) as \(v\) and using Leibnitz notation, this formula is often written as:

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