# Changes

## Integration by parts

, 04:45, 9 March 2016
Associated jargon
{{perspectives}}
{{integration rule}}
<section begin="statement"/>
! Version type !! Statement
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| Formal manipulation version for indefinite integration Indefinite integral in function notation || Suppose $F$ and $g$ are [[continuous function]]s such that $F$ is a [[differentiable function]] and we want to integrate $F(x)g(x)$. Suppose $G$ is an antiderivative for $g$ and $f = F'$. Then we have:<br>$\int F(x)g(x) \, dx = F(x)G(x) - \int f(x)G(x) \, dx$
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| Formal manipulation version for definite integration Definite integral in function notation || Suppose $F$ and $g$ are [[continuous function]]s on a [[closed interval]] $[a,b]$ such that $F$ is a [[differentiable function]] at all points of $[a,b]$ and we want to integrate $F(x)g(x)$. Suppose $G$ is an antiderivative for $g$ and $f$ is the derivative $F'$. Then, we have:<br>$\int_a^b F(x)g(x) \, dx = [F(x)G(x)]_a^b - \int_a^b f(x)G(x) \, dx$<br>The part $[F(x)G(x)]_a^b$ is shorthand for $F(b)G(b) - F(a)G(a)$, in keeping with the standard ''evaluate between limits'' notation used for definite integrals.
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| Formal manipulation version Indefinite integral in dependent-independent variable notation || Suppose $u,v$ are variables denoting functions of $x$. Then, we have:<br>$\int u \, dv = uv - \int v \, du$<br>More explicitly:<br>$\int u \frac{dv}{dx} \, dx = uv - \int v \frac{du}{dx} \, dx$<br>Compared with the notation of the preceding version, $u = F(x)$, $du/dx=f(x)$, $dv/dx = g(x)$, and $v = G(x)$.
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| Verbal version || The integral of a product of two functions is the first function times the integral of the second function minus the integral of (the derivative of the first function times the integral of the second function).
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===Associated jargon===

The function $F$ in the function notation (or the variable $u$ in in the variable notation) is termed the ''part to differentiate''. The function $g$ in the function notation (which is $dv/dx$ in the variable notation) is termed the ''part to integrate''.
===As a process===
We can think of integration by parts overall as a five- or six-step process. The really hard ''discretionary'' parts (i.e., the parts that are not purely procedural but require decision-making) are Steps (1) and (2):
*# Identify the function being integrated as a product of two functions*# Figure out which is the part to differentiate and which is the part to integrate*# Determine the antiderivative of the part to integrate and the derivative of the part to differentiate*# Apply integration by parts*# Now, do the second integral*# (Optional) Verify your answer by differentiating the antiderivative obtained
===Basic procedural observations===
* {| class="sortable" border="1"! Observation !! More information|-| Integration by parts ''essentially'' reverses the [[product rule for differentiation]] applied to $F(x)G(x)$ (or $uv$). See the || [[#proof|proof section]] for details.* |-| Solving a problem through a ''single'' application of integration by parts usually involves ''two'' integrations -- one to find the antiderivative $G$ for $g$ (which in the $u-v$ notation is equivalent to finding $v$ given $dv/dx$) and then doing the right side integration of $f(x)G(x)$ (or $v\frac{du}{dx}$). Note that:** <br>(1) For the first integration (the one involving finding $G$ or $v$), we just need to find ''one'' antiderivative and we ''do not'' put a $+C$ or try to evaluate between limits. See the section [[#Choice of antiderivative does not matter|choice of antiderivative does not matter]].** <br>(2) On the other hand, at the end of the second integration (the integration of $f(x)G(x)$ or $v\frac{du}{dx}$), we ''do'' put a $+C$ (for indefinite integration) or evaluate between limits (for definite integration).|| See the section [[#Choice of antiderivative does not matter|choice of antiderivative does not matter]]. For more details, see [[choice of antiderivative does not matter for integration by parts]]* |-| The key way that the original integral $\int F(x)g(x) \, dx$ (or $\int u \frac{dv}{dx} \, dx$) and the new integral $\int f(x)G(x) \, dx$ (or $\int v \frac{du}{dx} \, dx$) differ is that in the new integral, ''one part has been differentiated and the other part has been integrated''. For more, see the || The section [[#Equivalence of integration problems|equivalence of integration problems]].For more detailed information, see [[equivalence of integration problems arising from integration by parts]]|}
<section end="statement"/>
| One-sided differentiability at endpoints || At the endpoints, it is sufficient to require the appropriate one-sided differentiability and take the corresponding one-sided derivative. In symbols, if $F$ is differentiable on $(a,b)$, right differentiable at $a$, and left differentiable at $b$, we can take $f$ to equal $F'$ on $(a,b)$, the right hand derivative at $a$, and the left hand derivative at $b$.
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| Improper integrals || Integration by parts also works for improper integrals, ''provided'' the approriate appropriate limits make sense to compute. Improper integrals include the situation where one or both the limits of integration is infinite, or where the function is infinite or undefined at one or both of the endpoints of integration or somewhere within the domain of integration.
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Also, for trigonometric products, check out [[integration of product of sinusoidal functions]].
==Choosing the parts to integrate and differentiate==
| more convoluted trigonometric functions (such as [[tan]], [[sec]], and other expressions that are non-polynomial rational functions if expressed purely in terms of sine and cosine). Analogously, non-polynomial rational functions of hyperbolic sine and hyperbolic cosine || differentiation may make the expressions more complicated, but they stay within the trigonometric and exponential domain || in some cases, integration keeps the function within the trigonometric or exponential domain. These are reasonable to integrate<br>Others have antiderivatives that involve logarithms (such as the [[tangent function]], [[secant function]], etc.) and are terrible to integrate.<br>In many cases, one integration is fine but ''repeated integration'' eventually leads to a situation where we land up with a logarithmic function or get something we don't know how to integrate.
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===General heuristic: ILATE rule===
It should be remembered that there is no real precedence ordering between inverse trigonometric and logarithmic functions, and there is no real precedence ordering between trigonometric and exponential functions. However, it may still be good to adopt a consistent rule of precedence between trigonometric and exponential functions in order to avoid the [[#circular trap|circular trap]].
==Standard strategies for products and composites==
! Product type !! Preliminary thoughts !! Strategy for this product type !! Examples
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| polynomial function times basic trigonometric or exponential function (such as [[sin]], [[cos]], [[exp]], [[cosh]], [[sinh]], or a polynomial in such expressions) || The polynomial function can be differentiated, and repeated differentiation keeps making it simpler and simpler until it disappears. The sine or cosine function or exponential function, upon integration, does not get more complicated, and so can be integrated repeatedly. || Take the polynomial function as the part to differentiate, and keep using integration by parts repeatedly till the polynomial disappears. || Trig: $x \cos x, x \sin x, x^2 \cos x, x \tan^2 x$(see [[Practical:Integration by parts#Polynomial times basic trigonometric|here]] for worked out examples)<br>Exp: $xe^x, x^2e^x, x \cosh x, x\sinh x$(see [[Practical:Integration by parts#Polynomial times basic exponential|here]] for worked out examples)
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| inverse trigonometric function or logarithmic function (or composite of such a function with a polynomial function) times polynomial function (the ''polynomial function'' could just be the function $1$, which is invisible). || The polynomial function can easily be both differentiated and integrated. The inverse trigonometric or logarithmic function can be differentiated, bringing it into the algebraic domain. || Choose the inverse trigonometric function or logarithmic function as the part to differentiate. || Simple products:$\ln x, x \ln x, \arctan x, x \arctan x, \arcsin x$(see [[Practical:Integration by parts#Polynomial times logarithmic|here]] for worked out examples)<br>Products involving composites: $\ln(x^2 + 1), x\ln(x^3 + 1), x \arctan(x^3 - x)$
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| trigonometric function times exponential function OR product of trigonometric functions or power of a trigonometric function that can be treated as a product (and other techniques such as [[integration by u-substitution]] don't seem to solve the problem completely) || both functions are easy to differentiate and to integrate, with the complexity remaining the same. || We need to use the [[recursive version of integration by parts]]. || $\sin^2x, \sec^3x, e^x \cos x$<br>The function $\sec^3x$ can also be done using a similar method, though it falls outside the ''basic'' trigonometric function products.
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Here, we use integration by parts (one or more times) and combine that with some trigonometric identities or algebraic manipulations to see the original integrand re-appear unexpectedly. We then choose an antiderivative $I$ so that the linear equation holds without any additive constants, and solve the linear equation for $I$ to get the antiderivative $I$. For examples, see [[sec^3#Integration]] and [[sin^2#Integration]].
| inverse trigonometric function or logarithmic function (or composite of such a function with a polynomial function) times non-polynomial algebraic function ||The inverse trigonometric or logarithmic function, upon differentiation, lands in the algebraic domain. The hope is that the algebraic function can be integrated within the algebraic domain. || If the algebraic function can be integrated within the algebraic domain, then it works to pick the inverse trigonometric or logarithmic function as the part to differentiate, and the algebraic function as the part to integrate. || where it works: $(\arctan x)/x^2$ (because $1/x^2$ has an antiderivative $-1/x$ that is a rational function), $x \ln x/(x^2 + 1)^2$ (because $x/(x^2 + 1)^2$ has an antiderivative that is a rational function)<br>where it doesn't work: $(\arctan x)/x$ (because the antiderivative of $1/x$ is not a rational function), $(\ln x)/(x^2 + 1)$ (because the antiderivative of $1/(x^2 + 1)$ is not a rational function)