Integration by parts

In calculus, and more generally in mathematical analysis, integration by parts is a rule that transforms the integral of products of functions into other, possibly simpler, integrals. The rule arises from the product rule of derivative. Suppose ''f''(''x'') and ''g''(''x'') are two continuously differentiable functions. Then the integration by parts rule states that for endpoints ''a'', ''b'' :

\int_a^b f(x) g'(x)\,dx = \left[ f(x) g(x) \right]_{a}^{b} - \int_a^b  f'(x) g(x)\,dx

where we use the common notation :

\left[f(x) g(x) \right]_{a}^{b} = f(b) g(b) - f(a) g(a).

The rule is shown to be true by using the product rule for derivatives and the fundamental theorem of calculus. Thus :

f(b)g(b) - f(a)g(a) = \int_a^b \frac{d}{dx} (f(x) g(x)) \, dx =

\int_a^b f'(x) g(x) \, dx + \int_a^b f(x) g'(x) \, dx

In the traditional calculus curriculum, this rule is often stated using indefinite integrals in the form :

\int f(x) g'(x)\,dx = f(x) g(x) - \int g(x) f'(x)\,dx

or in an even shorter form, if we let ''u'' = ''f''(''x''), ''v'' = ''g''(''x'') and the differentials ''du'' = ''f''′(''x'') ''dx'' and ''dv'' = ''g''′(''x'') ''dx'', then it is in the form in which it is most often seen: :

\int u\,dv = u v - \int v\,du.

One can also formulate a discrete analogue for sequences, called summation by parts. Note that the original integral contains the derivative of ''g''; in order to be able to apply the rule, the antiderivative ''g'' must be found, and then the resulting integral ∫''g'' ''f''′ d''x'' must be evaluated. An alternative notation has the advantage that the factors of the original expression are identified as ''f'' and ''g'', but the drawback of a nested integral: :

\int f g\,dx = f \int g\,dx - \int \left ( f' \int g\,dx \right )dx

This formula is valid whenever ''f'' is continuously differentiable and ''g'' is continuous. == Application== The rule is helpful in the integration of a function ''h''(''x'') which can be expressed as a product of two functions, ''h''(''x'') = ''f''(''x'')''g''(''x''), in such a way that the derivative of ''f'' and an antiderivative of ''g'' are known, and the resulting integral of ''f'' ' times the integral of ''g'' can be evaluated. ==Examples== In order to calculate: :

\int x\cos (x) \,dx

Let: :''u'' = ''x'', so that ''du'' = ''dx'', :''dv'' = cos(''x'') ''dx'', so that ''v'' = sin(''x''). Then: :

\int x\cos (x) \,dx = \int u \,dv = uv - \int v \,du

\int x\cos (x) \,dx = x\sin (x) - \int \sin (x) \,dx

\int x\cos (x) \,dx = x\sin (x) + \cos (x) + C

where ''C'' is an arbitrary constant of integration. By repeatedly using integration by parts, integrals such as :

\int x^{3} \sin (x) \,dx \quad \mbox{and} \quad \int x^{2} e^{x} \,dx

can be computed in the same fashion: each application of the rule lowers the power of ''x'' by one. An interesting example that is commonly seen is: :

\int e^{x} \cos (x) \,dx

where, strangely enough, in the end, the actual integration does not need to be performed. This example uses integration by parts twice. First let: :''u'' = e^''x''; thus d''u'' = e^''x''d''x'' :''v'' = sin(''x''); thus d''v'' = cos(''x'')d''x'' Then: :

\int e^{x} \cos (x) \,dx = e^{x} \sin (x) - \int e^{x} \sin (x) \,dx

Now, to evaluate the remaining integral, we use integration by parts again, with: :''u'' = e^''x''; d''u'' = e^''x''d''x'' :''v'' = -cos(''x''); d''v'' = sin(''x'')d''x'' Then: :

\int e^{x} \sin (x) \,dx = -e^{x} \cos (x) - \int -e^{x} \cos (x) \,dx = -e^{x} \cos (x) + \int e^{x} \cos (x) \,dx

Putting these together, we get :

\int e^{x} \cos (x) \,dx = e^{x} \sin (x) + e^x \cos (x) - \int e^{x} \cos (x) \,dx

Notice that the same integral shows up on both sides of this equation. So we can simply add the integral to both sides to get: :

2 \int e^{x} \cos (x) \,dx = e^{x} \left( \sin (x) + \cos (x) \right)

\int e^{x} \cos (x) \,dx = {e^{x} \left( \sin (x) + \cos (x) \right) \over 2}

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 ''x'' is also known. The first example is ∫ ln(''x'') d''x''. We write this as: :

\int \ln (x) \cdot 1 \,dx

Let: :''u'' = ln(''x''); d''u'' = 1/''x'' d''x'' :''v'' = ''x''; d''v'' = 1·d''x'' Then: :

\int \ln (x) \,dx = x \ln (x) - \int x/x \,dx = x \ln (x) - \int 1 \,dx

\int \ln (x) \,dx = x \ln (x) - {x} + {C}

\int \ln (x) \,dx = x \left( \ln (x) - 1 \right) + C

where, again, C is the arbitrary constant of integration The second example is ∫ arctan(''x'') d''x'', where arctan(''x'') is the inverse tangent function. Re-write this as: :

\int 1 \cdot \arctan (x) \,dx

Now let: :''u'' = arctan(''x''); d''u'' = 1/(1+''x''²) d''x'' :''v'' = ''x''; d''v'' = 1·d''x'' Then: :

\int \arctan (x) \,dx = x \arctan (x) - \int {x \over (1 + x^2)} \,dx = x \arctan (x) - {1 \over 2} \log(1 + x^2) + C

using a combination of the inverse chain rule method and the natural logarithm integral condition. == Justification of the rule == Integration by parts follows from the product rule of differentiation: If the two continuously differentiable functions ''u''(''x'') and ''v''(''x'') are given, the product rule states that :

(uv)' = (uv' + vu')

By integrating both sides, we get :

uv = \int (uv' + vu') \,dx

The latter integral can be written as the sum of two integrals since integration is linear: :

uv = \int uv' \,dx + \int vu' \,dx

(the fact that ''u'' and ''v'' are ''continuously'' differentiable ensures that the two individual integrals exist.) Subtracting ∫ ''v''u′ d''x'' from both sides yields the desired formula of integration by parts. :

\int uv' \,dx = uv - \int vu' \,dx

Calculus Integral calculus

Integration by parts

:''An alternative notation has the advantage that the factors of the original expression are identified as f and g'' Why is that an advantage? It seems arbitrary to say it's better to call them ''f'' and ''g'' than to call them ''u'' and ''v''. User:Michael Hardy 00:27, 15 Nov 2003 (UTC) : That's not what I meant. The "classic" notation tells you what to do when you are integrating "f(x)g'(x)". The "alternative" integrates "f(x)g(x)". See? -- User:Tarquin 18:58, 15 Nov 2003 (UTC) ---- Could we chop of the ''justification'' section, since the rule is now justified at the beginning? Also I put the definite integral notation first, because the semantics of the indefinite integral form are a lot less clear. It is still in there since it may still be part of the calculus curriculum. Need to explain bound and free variables. The distribution comment at the end is informative. Maybe more can be said. The distribution article I think should also be rewritten, having two parts * a 1-dimensional dsitribution theory and * a general theory for open sets and manifolds. User:CSTAR 21:15, 11 May 2004 (UTC) == A practical note == ''I have removed this section from the end of the article. I think it probably belongs on Wikibooks, being a suggestion for students in second-semester calculus. I also did my best to clean up the writing style somewhat, to eliminate use of second person ("you") and incorrect word usage (e.g. the word "derivate", which may or may not actually mean "derivative", but if it does, it's not standard by any means) - please see my changes here: [http://en.wikipedia.org/w/index.php?title=Integration_by_parts&diff=0&oldid=8757154]. User:AerionUser_talk:Aerion 22:07, 23 Dec 2004 (UTC)'' In a general way, when an exponential or trigonometric function appears in the expression, it should be chosen as

v'

: :

\int f(x) e^{2x}\,dx = \frac{1}{2}  f(x) e^{2x} -  \frac{1}{2} \int f'(x) e^{2x}\,dx\ \dots

\int f(x) \cos(4x)\,dx = \frac{1}{4} f(x) \sin(4x) - \frac{1}{4} \int f'(x) \sin(4x)\,dx\ \dots

A good way to choose U and dV is the mnemonic "DETAIL". D stands for differential, which is what is to be chosen (dV). The rest of the letters give the order of functions to consider. E stands for exponential. T is for trigonometric function. A stands for algebraic function. I is for inverse trigonometric, and finally L is for logarithmic, which is typically a poor choice. This is especially useful when

f(x)

is a polynomial, since each consecutive derivative of f(x) is simpler, and eventually, is constant. In general, integration by parts if

f(x)

is easy to differentiate. Otherwise, the substitution rule may need to be employed. By contrast, logarithmic or inverse trigonometric functions should be chosen as

u

: :

\int f(x) \log(3x)\,dx = F(x) \log(3x) - \int F(x) \frac{3}{3x}\,dx\ \dots

\int f(x) \arcsin(6x)\,dx = F(x) \arcsin(6x) - \int F(x) \frac{6}{\sqrt{1 - (6x)^2}}\,dx\ \dots

The objective is to reduce the inverse trigonometric function to a fraction inside the integral. If the derivative contains a radical, trigonometric substitution may be useful. === ILATE rule === ''I have removed the text below, which is similar to the section I previously removed, seen above. It doesn't really fit the style of the rest of the article, and ought to be on Wikibooks instead, especially since it is clearly targeted for calculus students. :(User:AerionUser_talk:Aerion 23:03, 5 Feb 2005 (UTC)' Alternately a bit handy rule is ILATE rule. This rule helps to decide which function must be used as a substitute for ''f'' and which for ''g''. This rule works fine in most cases, making the calculations easier.
KEY:
I = Inverse functions
L = Logarithamic functions
A = Algebric functions
T = Trigonometric functions
E = Exponential functions
So if you get cos(''x'') and log(''x'') in the product then, according to the rule take "log(''x'')" as equivalent to "''f''" in the equation, while "cos(''x'')" takes the position of second function.
P.S. Sometimes you need to fiddle around and use the LIATE instead in some cases.

See other meanings of words starting from letter:

I

IA | IB | IC | ID | IE | IF | IG | IH | IJ | IK | IL | IM | IN | IO | IP | IR | IS | IT | IU | IW | IX | IY | IZ |

Words begining with Integration_by_parts:

Integration_by_parts
Integration_by_parts