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integration by parts
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Definition/Summary
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| In this article, we shall learn a method for integrating the product of two functions. This method is derived from the 'product rule' for differentiation, but can only be applied to integrate products of certain types. |
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Equations
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[tex]\int u dv=uv - \int v du[/tex]
where u and v are functions of one variable; x, say. |
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Recent forum threads on integration by parts
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Breakdown
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Mathematics
> Calculus/Analysis
>> Calculus
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Extended explanation
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As, you can see in the equation, it contains two variable, namely 'u' and 'v'. These variables are actually the representation of two functions and thus, the above rule can also be stated as:
[tex]\int f(x) \ g(x) \ dx=~ f(x)\int g(x) \ dx
\ -~\int \left[ \ f'(x) \int g(x) \ dx \ \right] \ dx
[/tex]
The most important step of initiating such problems would be the determination of u and v from the given function. This can be done by using the following order:
L- Logarithmic
I- Inverse trigonometric
A- Algebraic
T- Trigonometric
E- Exponential
(Or can be remembered as 'LIATE")
Thus, out of the two given function, whichever comes first in the above given list can considered as 'u', and hence the other to be 'v'.
Implementation of the rule given above can be clearly understood by solving an example. Consider,
[tex] I \equiv \int x cos(x) dx[/tex]
In the above stated example, 'x' is an algebraic expression whereas 'cos(x)' is a trigonometric expression. Thus, 'x' is considered as 'u' and 'cos(x)' is considered as 'v'. Hence, applying the above given rule, that is,
[tex]\int uv dx=~ u\int v dx-~\frac{du}{dx} \int vdx
[/tex]
so that
[tex]I= x\int cos(x) dx - \int\left[\frac{d}{dx}x\int cos(x) dx\right] dx[/tex]
[tex]= x \ sin(x) - \int sin(x) \ 1 \ dx +c_1 [/tex]
Because,
[tex]\int cos(x) dx= sin(x)~\text{and}~ \frac{d}{dx}x=~1[/tex]
And finally,
[tex]I = x \ sin(x) + cos(x) + c[/tex]
because
[tex]\int sin(x) dx= -cos(x)[/tex] |
Commentary
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