Editing Integration by parts (section)

===Finding antiderivatives===
Integration by parts is a [[heuristic]] rather than a purely mechanical process for solving integrals; given a single function to integrate, the typical strategy is to carefully separate this single function into a product of two functions ''u''(''x'')''v''(''x'') such that the residual integral from the integration by parts formula is easier to evaluate than the single function. The following form is useful in illustrating the best strategy to take:

<math display="block">\int uv\,dx = u \int v\,dx - \int\left(u' \int v\,dx \right)\,dx.</math>

On the right-hand side, ''u'' is differentiated and ''v'' is integrated; consequently it is useful to choose ''u'' as a function that simplifies when differentiated, or to choose ''v'' as a function that simplifies when integrated. As a simple example, consider:

<math display="block">\int\frac{\ln(x)}{x^2}\,dx\,.</math>

Since the derivative of ln(''x'') is {{sfrac|1|''x''}}, one makes (ln(''x'')) part ''u''; since the antiderivative of {{sfrac|1|''x''<sup>2</sup>}} is −{{sfrac|1|''x''}}, one makes {{sfrac|1|''x''<sup>2</sup>}} part ''v''. The formula now yields:

<math display="block">\int\frac{\ln(x)}{x^2}\,dx = -\frac{\ln(x)}{x} - \int \biggl(\frac1{x}\biggr) \biggl(-\frac1{x}\biggr)\,dx\,.</math>

The antiderivative of −{{sfrac|1|''x''<sup>2</sup>}} can be found with the [[power rule]] and is {{sfrac|1|''x''}}.

Alternatively, one may choose ''u'' and ''v'' such that the product ''u''′ (∫''v''&nbsp;''dx'') simplifies due to cancellation. For example, suppose one wishes to integrate:

<math display="block">\int\sec^2(x)\cdot\ln\Big(\bigl|\sin(x)\bigr|\Big)\,dx.</math>

If we choose ''u''(''x'') = ln(|sin(''x'')|) and ''v''(''x'') = sec<sup>2</sup>x, then ''u'' differentiates to <math>\frac{1}{\tan x}</math> using the [[chain rule]] and ''v'' integrates to tan ''x''; so the formula gives:

<math display="block">\int\sec^2(x)\cdot\ln\Big(\bigl|\sin(x)\bigr|\Big)\,dx = \tan(x)\cdot\ln\Big(\bigl|\sin(x)\bigr|\Big)-\int\tan(x)\cdot\frac1{\tan(x)} \, dx\ .</math>

The integrand simplifies to 1, so the antiderivative is ''x''. Finding a simplifying combination frequently involves experimentation.

In some applications, it may not be necessary to ensure that the integral produced by integration by parts has a simple form; for example, in [[numerical analysis]], it may suffice that it has small magnitude and so contributes only a small error term. Some other special techniques are demonstrated in the examples below.

====Polynomials and trigonometric functions====

In order to calculate

<math display="block">I=\int x\cos(x)\,dx\,,</math>

let:
<math display="block">\begin{alignat}{3}
   u &= x\ &\Rightarrow\ &&du &= dx \\
  dv &= \cos(x)\,dx\ &\Rightarrow\ && v &= \int\cos(x)\,dx = \sin(x)
\end{alignat}</math>

then:

<math display="block">\begin{align}
  \int x\cos(x)\,dx & = \int u\ dv \\
  & = u\cdot v - \int v \, du \\
  & = x\sin(x) - \int \sin(x)\,dx \\
  & = x\sin(x) + \cos(x) + C,
 \end{align}</math>

where ''C'' is a [[constant of integration]].

For higher powers of <math>x</math> in the form

<math display="block">\int x^n e^x\,dx,\ \int x^n\sin(x)\,dx,\ \int x^n\cos(x)\,dx\,,</math>

repeatedly using integration by parts can evaluate integrals such as these; each application of the theorem lowers the power of <math>x</math> by one.

====Exponentials and trigonometric functions====
{{hatnote|See also: [[Integration using Euler's formula]]}}
An example commonly used to examine the workings of integration by parts is

<math display="block">I=\int e^x\cos(x)\,dx.</math>

Here, integration by parts is performed twice. First let

<math display="block">\begin{alignat}{3}
   u &= \cos(x)\ &\Rightarrow\ &&du &= -\sin(x)\,dx \\
  dv &= e^x\,dx\ &\Rightarrow\ &&v  &= \int e^x\,dx = e^x
\end{alignat}</math>

then:

<math display="block">\int e^x\cos(x)\,dx = e^x\cos(x) + \int e^x\sin(x)\,dx.</math>

Now, to evaluate the remaining integral, we use integration by parts again, with:

<math display="block">\begin{alignat}{3}
   u &= \sin(x)\ &\Rightarrow\ &&du &= \cos(x)\,dx \\
  dv &= e^x\,dx\,&\Rightarrow\ && v &= \int e^x\,dx = e^x.
\end{alignat}</math>

Then:

<math display="block">\int e^x\sin(x)\,dx = e^x\sin(x) - \int e^x\cos(x)\,dx.</math>

Putting these together,

<math display="block">\int e^x\cos(x)\,dx = e^x\cos(x) + e^x\sin(x) - \int e^x\cos(x)\,dx.</math>

The same integral shows up on both sides of this equation. The integral can simply be added to both sides to get

<math display="block">2\int e^x\cos(x)\,dx = e^x\bigl[\sin(x)+\cos(x)\bigr] + C,</math>

which rearranges to

<math display="block">\int e^x\cos(x)\,dx = \frac{1}{2}e^x\bigl[\sin(x)+\cos(x)\bigr] + C'</math>

where again <math>C</math> (and <math>C' = \frac{C}{2}</math>) is a [[constant of integration]].

A similar method is used to find the [[integral of secant cubed]].

====Functions multiplied by unity====

Two other well-known examples are when integration by parts is applied to a function expressed as a product of 1 and itself. This works if the derivative of the function is known, and the integral of this derivative times <math>x</math> is also known.

The first example is <math>\int \ln(x) dx</math>. We write this as:

<math display="block">I=\int\ln(x)\cdot 1\,dx\,.</math>

Let:

<math display="block">u = \ln(x)\ \Rightarrow\ du = \frac{dx}{x}</math>
<math display="block">dv = dx\ \Rightarrow\ v = x</math>

then:

<math display="block">
\begin{align}
\int \ln(x)\,dx & = x\ln(x) - \int\frac{x}{x}\,dx \\
& = x\ln(x) - \int 1\,dx \\
& = x\ln(x) - x + C
\end{align}
</math>

where <math>C</math> is the [[constant of integration]].

The second example is the [[inverse tangent]] function <math>\arctan(x)</math>:

<math display="block">I=\int\arctan(x)\,dx.</math>

Rewrite this as

<math display="block">\int\arctan(x)\cdot 1\,dx.</math>

Now let:

<math display="block">u = \arctan(x)\ \Rightarrow\ du = \frac{dx}{1+x^2}</math>

<math display="block">dv = dx\ \Rightarrow\ v = x</math>

then

<math display="block">
\begin{align}
\int\arctan(x)\,dx
& = x\arctan(x) - \int\frac{x}{1+x^2}\,dx \\[8pt]
& = x\arctan(x) - \frac{\ln(1+x^2)}{2} + C
\end{align}
</math>

using a combination of the [[inverse chain rule method]] and the [[natural logarithm integral condition]].

====LIATE rule====
The LIATE rule is a rule of thumb for integration by parts. It involves choosing as ''u'' the function that comes first in the following list:<ref>{{Cite journal |jstor=2975556 |first=Herbert E. |last=Kasube |title=A Technique for Integration by Parts |journal=[[The American Mathematical Monthly]] |volume=90 |issue=3 |year=1983 |pages=210–211 |doi=10.2307/2975556}}</ref>
* '''L''' – [[logarithmic function]]s: <math>\ln(x),\ \log_b(x),</math> etc.
* '''I''' – [[inverse trigonometric function]]s (including [[Inverse hyperbolic functions|hyperbolic analogues]]):  <math>\arctan(x),\ \arcsec(x),\ \operatorname{arsinh}(x),</math> etc.
* '''A''' – [[algebraic function]]s (such as [[polynomials]]): <math>x^2,\ 3x^{50},</math> etc.
* '''T''' – [[trigonometric functions]] (including [[Hyperbolic functions|hyperbolic analogues]]): <math>\sin(x),\ \tan(x),\ \operatorname{sech}(x),</math> etc.
* '''E''' – [[exponential function]]s: <math>e^x,\ 19^x,</math> etc.

The function which is to be ''dv'' is whichever comes last in the list. The reason is that functions lower on the list generally have simpler [[antiderivative]]s than the functions above them. The rule is sometimes written as "DETAIL", where ''D'' stands for ''dv'' and the top of the list is the function chosen to be ''dv''. An alternative to this rule is the ILATE rule, where inverse trigonometric functions come before logarithmic functions.

To demonstrate the LIATE rule, consider the integral

<math display="block">\int x \cdot \cos(x) \,dx.</math>

Following the LIATE rule, ''u'' = ''x'', and ''dv'' = cos(''x'') ''dx'', hence ''du'' = ''dx'', and ''v'' = sin(''x''), which makes the integral become
<math display="block">x \cdot \sin(x) - \int 1 \sin(x) \,dx,</math>
which equals
<math display="block">x \cdot \sin(x) + \cos(x) + C.</math>

In general, one tries to choose ''u'' and ''dv'' such that ''du'' is simpler than ''u'' and ''dv'' is easy to integrate. If instead cos(''x'') was chosen as ''u'', and ''x dx'' as ''dv'', we would have the integral

<math display="block">\frac{x^2}{2} \cos(x) + \int \frac{x^2}{2} \sin(x) \,dx,</math>

which, after recursive application of the integration by parts formula, would clearly result in an infinite recursion and lead nowhere.

Although a useful rule of thumb, there are exceptions to the LIATE rule. A common alternative is to consider the rules in the "ILATE" order instead. Also, in some cases, polynomial terms need to be split in non-trivial ways. For example, to integrate

<math display="block">\int x^3 e^{x^2} \,dx,</math>

one would set

<math display="block">u = x^2, \quad dv = x \cdot e^{x^2} \,dx,</math>

so that

<math display="block">du = 2x \,dx, \quad v = \frac{e^{x^2}}{2}.</math>

Then

<math display="block">\int x^3 e^{x^2} \,dx = \int \left(x^2\right) \left(xe^{x^2}\right) \,dx = \int u \,dv
= uv - \int v \,du = \frac{x^2 e^{x^2}}{2} - \int x e^{x^2} \,dx.</math>

Finally, this results in
<math display="block">\int x^3 e^{x^2} \,dx = \frac{e^{x^2}\left(x^2 - 1\right)}{2} + C.</math>

Integration by parts is often used as a tool to prove theorems in [[mathematical analysis]].