4. Integration by Parts

Recall:       \(\displaystyle \int u\,dv=u\,v-\int v\,du\)   where   \(du=\dfrac{du}{dx}\,dx\)   and   \(dv=\dfrac{dv}{dx}\,dx\)

b1. Choosing the 'Parts'

In applying the integration by parts formula, the difficult part is deciding what to pick as \(u\) and as \(dv\). Here are some guidelines.

General Rules:

• Pick \(u\) as a function which is easy to differentiate.
• Pick \(dv\) as a function which is easy to integrate.
• Pick \(u\) and \(dv\) so that \(v\,du\) is simpler than \(u\,dv.\)

Special Rules:

• Polynomials are easy to differentiate or integrate, but the derivative is simpler. So it is a good idea to take a polynomial for \(u\).
• The functions \(e^x\), \(\sin x\) and \(\cos x\) are easy to differentiate or integrate and don't get more complicated. So it is a good idea to take these functions for \(dv\).
• The functions \(\ln x\), \(\arcsin x\) and \(\arctan x\) are easy to differentiate but hard to integrate. So it is a good idea to take these functions for \(u\).

These rules are summarized in the following:

LAPTE Mnemonic for Selecting \(u\)

L:	\(\ln\) or \(\log\) factor
A:	\(\arcsin\) or \(\arctan\) factor
P:	polynomial factor
T:	trigonometric factor
E:	exponential factor

\(\displaystyle \int x\sec^2x\,dx\)

Since \(x\) is a polynomial (the “P” in LAPTE.) and \(\sec^2 x\) can be easily integrated, we take the parts: \[\begin{array}{ll} u=x & dv=\sec^2x\,dx \\ du=dx \quad & v=\tan x \end{array}\] This gives \[\begin{aligned} \displaystyle\int x\sec^2x\,dx &=x\tan x-\int \tan x\,dx \\ &=x\tan x-\int \dfrac{\sin x}{\cos x}\,dx \end{aligned}\] We now use the substitution \(u=\cos x\), with \(du=-\sin x\,dx\) to get \[\begin{aligned} \displaystyle \int x\sec^2x\,dx &=x\tan x+\int \dfrac{1}{u}\,du \\ &=x\tan x+\ln|u|+C \\ &=x\tan x+\ln|\cos x|+C \end{aligned}\]

We check by differentiating. If   \(f(x)=x\tan x+\ln|\cos x|\),   then \[ f'(x)=\tan x+x\sec^2 x+\dfrac{-\sin x}{\cos x}=x\sec^2 x \]

Here are some more examples:

Polynomial Examples

Log Examples

Inverse Trig Examples

We wish to compute the following integrals using integration by parts. What 'parts' should be selected

1. \(\displaystyle \int x\sin 2x\,dx\)

   \(\displaystyle\begin{aligned}&u=\sin 2x \\ &dv=x\,dx\end{aligned}\)

   \(\displaystyle\begin{aligned}&u=2x \\ &dv=\sin x\,dx\end{aligned}\)

   \(\displaystyle\begin{aligned}&u=x\sin 2x \\ &dv=dx\end{aligned}\)

   \(\displaystyle\begin{aligned}&u=x \\ &dv=\sin 2x\,dx\end{aligned}\)

   A. Incorrect. This selection of parts gives \(du=2\cos 2x\,dx\) and \(v=\dfrac{1}{2}x^2\). This gives \[ \int x\sin 2x\,dx=\dfrac{1}{2}x^2\sin 2x-\int x^2\sin 2xdx \] The problem is more complex than before.

   B. Sorry. There is no expression \(\sin x\) in this problem. Read the problem carefully. Try again.

   C. Incorrect. Beginning with \(u=x\sin 2x\)and \(dv=dx\), we have \(du=\left(\sin 2x+2x\cos 2x\right)\,dx\) and \(v=x\). Then: \[\int x\sin 2x\,dx=x^2\sin 2x-\int x\left(\sin 2x+2x\cos 2x\right)\,dx\] This is far more complex than the original integral.

   D. Correct. With these parts, \(du=dx\) and \(v=-\,\dfrac{1}{2}\cos 2x\). So: \[\begin{aligned} \int x\sin 2x\,dx&=-\,\dfrac{1}{2}x\cos 2x+\dfrac{1}{2}\int \cos 2x\,dx \\ &=-\,\dfrac{1}{2}x\cos 2x+\dfrac{1}{4}\sin 2x+C \end{aligned}\]

2. \(\displaystyle \int x^3 \ln 4x\,dx\)

   \(\displaystyle\begin{aligned}&u=x^3 \\ &dv=\ln 4x\,dx\end{aligned}\)

   \(\displaystyle\begin{aligned}&u=\ln 4x \\ &dv=x^3\,dx\end{aligned}\)

   \(\displaystyle\begin{aligned}&u=\ln 4x \\ &dv=x^3\end{aligned}\)

   \(\displaystyle\begin{aligned}&u=\ln 4x \\ &dv=x\,dx\end{aligned}\)

   A. Sorry. This doesn't follow the order given by LAPTE. We don't want to include \(\displaystyle \ln 4x\) in the \(\displaystyle dv\) expression because it is not easy to take its antiderivative.

   B. Correct. With these parts we get \(\displaystyle du=\dfrac{1}{x}\,dx\) and \(v=\dfrac{x^4}{4}\). So: \[\begin{aligned} \int x^3\ln 4x\,dx &=\dfrac{x^4}{4}\ln 4x-\dfrac{1}{4}\int x^3\,dx \\ &=\dfrac{x^4}{4}\ln 4x-\dfrac{1}{16}x^4+C \end{aligned}\]

   C. Sorry. The \(dx\) needs to go in the \(dv\) expression.

   D. Incorrect. After choosing the parts, there is still an \(x^2\) left in the integrand.

3. \(\displaystyle \int (4x^3+2x)\arctan x\,dx\)

   \(\displaystyle\begin{aligned}&u=4x^3 \\ &dv=\arctan x\end{aligned}\)

   \(\displaystyle\begin{aligned}&u=2x \\ &dv=\arctan x\,dx\end{aligned}\)

   \(\displaystyle\begin{aligned}&u=4x^3+2x \\ &dv=\arctan x\,dx\end{aligned}\)

   \(\displaystyle\begin{aligned}&u=\arctan x \\ &dv=(4x^3+2x)\,dx\end{aligned}\)

   A. Sorry that's not right. It looks like you forgot to include \(\displaystyle dx\).

   B. It's close, but it's not quite right. You've forgotten about the \(\displaystyle 4x^3\).

   C. While this answer is technically correct, it will be significantly harder to integrate \(\displaystyle dv=\arctan x\,dx\). Pay attention to LAPTE.

   D. Correct. After choosing these parts, we get \(\displaystyle du=\dfrac{1}{1+x^2}\,dx\) and \(\displaystyle v=x^4+x^2\). So: \[\begin{aligned} \int (4x^3+2x)\arctan x\,dx &=(x^4+x^2)\arctan x-\int\dfrac{x^4+x^2 }{1+x^2}\,dx \\ &=(x^4+x^2)\arctan x-\int \dfrac{x^2(x^2+1)}{1+x^2}\,dx \\ &=(x^4+x^2)\arctan x-\int x^2\,dx \\ &=(x^4+x^2)\arctan x-\dfrac{x^3}{3}+C \end{aligned}\]

You can also practice computing indefinite integrals using Integration by Parts by using the following Maplet (requires Maple on the computer where this is executed):

Integration by Parts   Rate It

At this point, you should only click on the Indefinite Integral button.