2.4 Integration By Parts

We have already seen in Section 2.2.2 that recognizing the Product Rule can be useful, when we noticed that

\begin{equation*} \int \sec^3u+\sec u \tan^2u\,du=\sec u \tan u\text{.} \end{equation*}

As with substitution, we do not have to rely on insight or cleverness to discover such antiderivatives; there is a technique that will often help to uncover the Product Rule called Integration by Parts.

Start with the Product Rule:

\begin{equation*} {d\over dx}f(x)g(x)=f'(x)g(x)+f(x)g'(x)\text{.} \end{equation*}

We can apply integration to this equation and obtain

\begin{equation*} f(x)g(x)=\int f'(x)g(x)\,dx +\int f(x)g'(x)\,dx\text{,} \end{equation*}

and then rewrite this as

\begin{equation*} \int f(x)g'(x)\,dx=f(x)g(x)-\int f'(x)g(x)\,dx\text{.} \end{equation*}

This may not seem particularly useful at first glance, but it turns out that in many cases we have an integral of the form

\begin{equation*} \int f(x)g'(x)\,dx \end{equation*}

but that

\begin{equation*} \int f'(x)g(x)\,dx \end{equation*}

is easier to integrate.

This technique for turning one integral into another is called Integration by Parts, and is usually written in more compact form.

Theorem 2.31. Integration by Parts.

Let \(u\) and \(v\) be differentiable functions, then

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

where

\begin{equation*} u=f(x) \text{ and } v=g(x) \text{ so that } du=f'(x)\,dx \text{ and } dv=g'(x)\,dx\text{.} \end{equation*}

Note:

1. To use this technique we need to identify likely candidates for \(u=f(x)\) and \(dv=g'(x)\,dx\text{.}\) When choosing \(u\) and \(dv\text{,}\) keep in mind that we need to be able to readily find an antiderivative for \(dv\) and that \(du\) becomes simpler than \(u\text{.}\) Simpler could mean the power is reduced by one degree, or the original integral appears on the right side, or ...

2. After we have applied Integration by Parts, we then need to integrate \(\ds \int v\,du\text{.}\) There is a danger to fall into a circular trap by choosing as the part to integrate (\(v\)) the term in the differential (\(du\)) from the first application of Integration by Parts. This does not provide you with any new information, but instead brings you back to the original integral. For example: To evaluate \(\ds \int x^2\sin x\,dx\) we choose

   \begin{equation*} u=x^2 \text{ and } v=\sin x\,dx \text{ so that } du=2x\,dx \text{ and } dv=-\cos x\,dx\text{,} \end{equation*}

   then

   \begin{equation*} \int x^2\sin x\,dx = -x^2\cos x + 2\int x\cos x\,dx\text{.} \end{equation*}

   If we ignore that the new integral is simpler than the original integral, which would tell us to continue in the same manner of selecting \(u\) and \(dv\text{,}\) we may fall into the circular trap of choosing

   \begin{equation*} u=\cos x \text{ and } v=x\,dx \text{ so that } du=-\sin x\,dx \text{ and } dv=\frac{x^2}{2}\,dx\text{,} \end{equation*}

   so that

   \begin{equation*} \begin{split} \int x^2\sin x\,dx \amp = -x^2\cos x + 2\int x\cos x\,dx \\ \amp = -x^2\cos x + 2\left(\cos x \frac{x^2}{2} + \int \frac{x^2}{2}\sin x\,dx\right) \\ \amp = \int x^2 \sin x\,dx. \end{split} \end{equation*}

   This shows that with our carelessness we have wasted our time and are back at the beginning.

Example 2.32. Product of a Linear Function and Logarithm.

Evaluate \(\ds\int x\ln x\,dx\text{.}\)

Solution

Let \(u=\ln x\) so \(du=1/x\,dx\text{.}\) Then we must let \(dv=x\,dx\) so \(\ds v=x^2/2\) and

\begin{equation*} \begin{split} \int x\ln x\,dx\amp={x^2\ln x\over 2}-\int {x^2\over2}{1\over x}\,dx\\ \amp= {x^2\ln x\over 2}-\int {x\over2}\,dx={x^2\ln x\over 2}-{x^2\over4}+C.\end{split} \end{equation*}

Example 2.33. Inverse Trigonometric Function.

Evaluate \(\ds\int \sin^{-1} x \, dx\text{.}\)

Solution

Let \(u=\sin^{-1} x\) so

\begin{equation*} du=\frac{1}{\sqrt{1-x^2}}\, dx\text{.} \end{equation*}

Then we must let \(dv=\,dx\) so \(v=x\text{.}\) Therefore,

\begin{equation*} \int \sin^{-1} x\,dx = x\sin^{-1} x-\int \frac{x}{\sqrt{1-x^2}}\,dx\text{.} \end{equation*}

Now use substitution with \(u = 1-x^2\) and \(du = -2x\, dx\text{.}\) Then

\begin{equation*} \begin{split} \int \frac{x}{\sqrt{1-x^2}}\,dx \amp = -\frac{1}{2}\int\frac{1}{\sqrt u}\,du\\ \amp =-\sqrt{u}+C\\ \amp =-\sqrt{1-x^2}+C. \end{split} \end{equation*}

Altogether, we find that

\begin{equation*} \int \sin^{-1} x\,dx = x\sin^{-1} x+\sqrt{1-x^2}+C\text{.} \end{equation*}

Example 2.34. Secant Cubed (again).

Evaluate \(\ds\int\sec^3 x\,dx\text{.}\)

Solution

Of course we already know the answer to this, but we needed to be clever to discover it. Here we'll use the new technique to discover the antiderivative. Let \(u=\sec x\) and \(\ds dv=\sec^2 x\,dx\text{.}\) Then \(du=\sec x\tan x\) and \(v=\tan x\) and

\begin{align*} \int\sec^3 x\,dx\amp = \sec x\tan x-\int \tan^2x\sec x\,dx\\ \amp = \sec x\tan x-\int (\sec^2x-1)\sec x\,dx\\ \amp = \sec x\tan x-\int \sec^3x\,dx +\int\sec x\,dx\text{.} \end{align*}

At first this looks useless—we're right back to \(\ds \int\sec^3x\,dx\text{.}\) But looking more closely, we notice that we can add this integral to both sides and are left to deal with the integral \(\ds \int \sec x\,dx\text{.}\)

\begin{align*} \int\sec^3x\,dx\amp = \sec x\tan x-\int \sec^3x\,dx +\int\sec x\,dx\\ \int\sec^3x\,dx+\int \sec^3x\,dx\amp = \sec x\tan x +\int\sec x\,dx\\ 2\int\sec^3x\,dx\amp = \sec x\tan x +\int\sec x\,dx\\ \int\sec^3x\,dx\amp = {\sec x\tan x\over2} +{1\over2}\int\sec x\,dx \end{align*}

Now we use our knowledge of \(\ds\int \sec x\,dx\) to conclude that

\begin{equation*} \int\sec^3 x\,dx = {\sec x\tan x\over2} +{\ln|\sec x+\tan x|\over2}+C\text{.} \end{equation*}

Example 2.35. Product of a Polynomial and Trigonometric Function.

Evaluate \(\ds\int x^2\sin x\,dx\text{.}\)

Solution

Let \(u=x^2\text{,}\) \(dv=\sin x\,dx\text{;}\) then \(du=2x\,dx\) and \(v=-\cos x\text{.}\) Now

\begin{equation*} \ds \int x^2\sin x\,dx=-x^2\cos x+\int 2x\cos x\,dx\text{.} \end{equation*}

This is better than the original integral since the power of \(x\) has been reduced by one degree, but we need to do Integration by Parts again. Let \(u=2x\text{,}\) \(dv=\cos x\,dx\text{;}\) then \(du=2\) and \(v=\sin x\text{,}\) and

\begin{align*} \int x^2\sin x\,dx\amp = -x^2\cos x+\int 2x\cos x\,dx\\ \amp = -x^2\cos x+ 2x\sin x - \int 2\sin x\,dx\\ \amp = -x^2\cos x+ 2x\sin x + 2\cos x + C\text{.} \end{align*}