Integration by Parts - HMC Calculus Tutorial

$back to the math tutorial index$

Integration by Parts

We will use the Product Rule for derivatives to derive a powerful integration formula:

Start with .
Integrate both sides to get $\displaystyle f(x)g(x)=\int\! f(x)g'(x)\, dx +\int\! f'(x)g(x)\, dx$ . (We need not include a constant of integration on the left, since the integrals on the right will also have integration constants.)
Solve for $\displaystyle\int\! f(x)g'(x)\, dx$ , obtaining

$\begin{displaymath}\int f(x)g'(x)\, dx=f(x)g(x)-\int f'(x)g(x)\, dx.\end{displaymath}$

This formula frequently allows us to compute a difficult integral by computing a much simpler integral. We often express the Integration by Parts formula as follows:

Let

$\begin{displaymath} \begin{array}{ll} u = f(x)\qquad\qquad & dv = g'(x)\, dx\\ du = f'(x)\, dx & v = g(x) \end{array}\end{displaymath}$

Then the formula becomes

$\begin{displaymath}\int u\, dv=uv-\int v\, du.\end{displaymath}$

To integrate by parts, strategically choose u, dv and then apply the formula.

Example

Let's evaluate $\displaystyle\int\! xe^x\, dx$ .

Let

$\begin{displaymath} \begin{array}{ll} u = x \qquad\qquad & dv = e^x\, dx\\ du = dx & v = e^x \end{array}\end{displaymath}$

Then by integration by parts,

$\begin{eqnarray*} \int xe^x&=&xe^x-\int e^x\, dx\\ &=&xe^x-e^x+C. \end{eqnarray*}$

A Faulty Choice A Reduction Formula

Integration by parts ``works'' on definite integrals as well:

$\begin{displaymath}\int^b_a u\, dv=\left.uv\right\vert^b_a-\int^b_a v\, du.\end{displaymath}$

Example

We will evaluate $\displaystyle\int^1_0\! \arctan (x)\, dx$ .

Let

$\begin{displaymath} \begin{array}{ll} u = \arctan(x) \qquad\qquad & dv = dx\\ [5pt] du = \displaystyle\frac{1}{1+x^2}\, dx & v = x \end{array}\end{displaymath}$

Then by integration by parts,

$\begin{eqnarray*} \int^1_0 \arctan(x)&=&\left.x\arctan(x)\right\vert^1_0-\int^1_... ...rac{1}{2} \ln (2)-0\right)\\ &=&\frac{\pi}{4}-\ln (\sqrt{2}). \end{eqnarray*}$

Sometimes it is necessary to integrate twice by parts in order to compute an integral:

Example

Let's compute $\displaystyle\int\! e^x\cos x\, dx$ .

Let

$\begin{displaymath} \begin{array}{ll} u = e^x\qquad\qquad & dv = \cos x\, dx\\ du = e^x\, dx & v = \sin x \end{array}\end{displaymath}$

Then $\displaystyle\int\! e^x\cos x\, dx=e^x\sin x-\int\! e^x\sin x\, dx$ .

It is not clear yet that we've accomplished anything, but now let's integrate the integral on the right-hand side by parts:

Now let

$\begin{displaymath} \begin{array}{ll} u = e^x\qquad\qquad & dv = \sin x\, dx\\ du = e^x\, dx & v = -\cos x \end{array}\end{displaymath}$

So $\displaystyle\int\! e^x\sin x\, dx=-e^x\cos x+\int e^x\cos x\, dx$ .

Substituting this into $\displaystyle\int\! e^x\cos x\, dx=e^x\sin x-\int\! e^x\sin x\, dx$ ,

$\begin{eqnarray*} \int e^x\cos x\, dx&=&e^x\sin x-\left[-e^x\cos x+\int e^x\cos x\, dx\right]\\ &=&e^x\sin x+ e^x\cos x -\int e^x\cos x\, dx. \end{eqnarray*}$

The integal $\displaystyle\int\! e^x\cos x\, dx$ appears on both sides on the equation, so we can solve for it:

$\begin{displaymath}2\int e^x\cos x\, dx=e^x\sin x+e^x\cos x.\end{displaymath}$

Finally,

$\begin{displaymath}\int e^x\cos x\, dx =\frac{1}{2}e^x\sin x+\frac{1}{2}e^x\cos x +C.\end{displaymath}$

Check by Differentiating

Key Concept

$\begin{displaymath}\int u\, dv=uv-\int v\, du.\end{displaymath}$

Choose u, dv in such a way that:
1. u is easy to differentiate.
2. dv is easy to integrate.
3. $\displaystyle\int\! v\, du$ is easier to compute that $\displaystyle\int\! u\, dv$ .
Sometimes it is necessary to integrate by parts more than once.

[I'm ready to take the quiz.] [I need to review more.]

[Take me back to the Tutorial Page]