Difference between revisions of "Integration by parts"
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This article details the method known as '''Integration by Parts'''. | This article details the method known as '''Integration by Parts'''. | ||
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This rule is often useful when one function is a power of ''x'' and the other function is a trigonometric function or ''e'' raised to a power of ''x''. | This rule is often useful when one function is a power of ''x'' and the other function is a trigonometric function or ''e'' raised to a power of ''x''. | ||
| − | Note that it may be necessary to repeat the '''integration by parts''' several times, | + | Note that it may be necessary to repeat the '''integration by parts''' several times, depending on the situation. |
===Proving Integration by Parts=== | ===Proving Integration by Parts=== | ||
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<br /> | <br /> | ||
If both sides are integrated: | If both sides are integrated: | ||
| − | :<math>f(x)g(x) = \int f'(x)g(x) + \int f(x)g'(x)</math> | + | :<math>f(x)g(x) = \int f'(x)g(x)dx + \int f(x)g'(x)dx</math> |
<br /> | <br /> | ||
By rearranging the terms: | By rearranging the terms: | ||
| − | :<math>\int f(x)g'(x) = f(x)g(x) - \int f'(x)g(x)</math> | + | :<math>\int f(x)g'(x)dx = f(x)g(x) - \int f'(x)g(x)dx</math> |
| + | Sometimes, Integration by Parts is simply expressed as: | ||
| + | :<math>\int udv = uv - \int vdu</math> | ||
| + | |||
| + | ==Using Integration by Parts== | ||
| + | |||
| + | === Example 1 === | ||
| + | A common example where integration by parts is needed is this: | ||
| + | :<math>\int x\sin(x)dx</math> | ||
| + | Let's set: | ||
| + | :<math>u = x</math> | ||
| + | :<math>dv = \sin(x)</math> | ||
| + | We can begin to solve: | ||
| + | :<math>du = dx</math> | ||
| + | :<math>v = -\cos(x)</math> | ||
| + | :<math>\int x\sin(x)dx = -x\cos(x) - \int -\cos(x) dx</math> | ||
| + | :<math>= -x\cos(x) + \int \cos(x) dx</math> | ||
| + | Now we are left with an integral that we '''do''' know how to solve: | ||
| + | :<math>= -x\cos(x) + sin(x) + C</math> | ||
| + | Therefore: | ||
| + | :<math>\int x\sin(x)dx = -x\cos(x) + sin(x) + C</math> | ||
| + | |||
| + | === Example 2 === | ||
| + | <br />Say that we are given: | ||
| + | :<math>\int e^x \sin(x)dx</math> | ||
| + | Let's set: | ||
| + | :<math>u=\sin(x)</math> | ||
| + | :<math>dv=e^xdx</math> | ||
| + | And continue solving: | ||
| + | :<math>du = \cos(x)dx</math> | ||
| + | :<math>v = e^x</math> | ||
| + | :<math>\int e^x \sin(x)dx</math> | ||
| + | And continue solving: | ||
| + | :<math>= \sin(x)e^x - \int e^x \cos(x) dx</math> | ||
| + | Now, we set: | ||
| + | :<math>u = \cos(x) </math> | ||
| + | :<math>dv = e^xdx </math> | ||
| + | And continue solving: | ||
| + | :<math>du = -\sin(x)dx</math> | ||
| + | :<math>v = e^x</math> | ||
| + | :<math>\sin(x)e^x - \int e^x \cos(x) dx</math> | ||
| + | :<math>= \sin(x)e^x - \cos(x)e^x + \int -e^x\sin(x) dx = \sin(x)e^x - \cos(x)e^x - \int e^x\sin(x) dx</math> | ||
| + | Let's look back at what we have done: | ||
| + | :<math>\int e^x \sin(x)dx = \sin(x)e^x - \cos(x)e^x - \int e^x\sin(x) dx</math> | ||
| + | The integral of the function has the integral of itself ... inside itself. This is actually a common occurrence with functions like the one we are solving. However, we can cheat the never-ending cycle of integrals by doing this: | ||
| + | :<math>2\int e^x \sin(x)dx = \sin(x)e^x - \cos(x)e^x</math> | ||
| + | :<math>\int e^x \sin(x)dx = {\sin(x)e^x - \cos(x)e^x \over 2}</math> | ||
==Rapid Repeated Integration== | ==Rapid Repeated Integration== | ||
'''Rapid Repeated Integration''' is a shortcut method for reduction problems that require Integration by Parts. It is especially useful when one function's derivative reduces to zero. | '''Rapid Repeated Integration''' is a shortcut method for reduction problems that require Integration by Parts. It is especially useful when one function's derivative reduces to zero. | ||
| + | |||
| + | ===Example 3=== | ||
| + | |||
For example, integrating: | For example, integrating: | ||
:<big><math>\int x^4 sin(x)\,dx</math></big> | :<big><math>\int x^4 sin(x)\,dx</math></big> | ||
| Line 38: | Line 87: | ||
|- | |- | ||
| <math>x^4</math> | | <math>x^4</math> | ||
| − | | <math>sin(x)</math> | + | | <math>\sin(x)</math> |
|- | |- | ||
| <math>4x^3</math> | | <math>4x^3</math> | ||
| − | | <math>-cos(x)</math> | + | | <math>-\cos(x)</math> |
|- | |- | ||
| <math>12x^2</math> | | <math>12x^2</math> | ||
| − | | <math>-sin(x)</math> | + | | <math>-\sin(x)</math> |
|- | |- | ||
| <math>24x</math> | | <math>24x</math> | ||
| − | | <math>cos(x)</math> | + | | <math>\cos(x)</math> |
|- | |- | ||
| <math>24</math> | | <math>24</math> | ||
| − | | <math>sin(x)</math> | + | | <math>\sin(x)</math> |
|- | |- | ||
| <math>0</math> | | <math>0</math> | ||
| − | | <math>-cos(x)</math> | + | | <math>-\cos(x)</math> |
|} | |} | ||
| Line 63: | Line 112: | ||
So then the integral becomes: | So then the integral becomes: | ||
| − | :<big><math>\int x^4 sin(x)\,dx</math></big> = <big><math>\ -x^4cos(x)-(-4x^3sin(x))+(12x^4cos(x))-(24xsin(x))-(-24cos(x))</math></big>=<big><math>\ -x^4cos(x)+4x^3sin(x)+12x^2cos(x)-24xsin(x)-24cos(x)+ | + | :<big><math>\int x^4 sin(x)\,dx</math></big> = <big><math>\ -x^4cos(x)-(-4x^3sin(x))+(12x^4cos(x))-(24xsin(x))-(-24cos(x))</math> |
| + | :</big>=<big><math>\ -x^4cos(x)+4x^3sin(x)+12x^2cos(x)-24xsin(x)-24cos(x)+C</math></big> | ||
| + | |||
| + | While this method is very useful when part of the integrand becomes zero through [[Derivative (calculus)|derivation]], it does not work in cases such as Example 2 above. | ||
| − | == See | + | == See also == |
*[[Integration]] | *[[Integration]] | ||
*[[Methods of integration]] | *[[Methods of integration]] | ||
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{{reflist}} | {{reflist}} | ||
| − | [[ | + | [[Category:Calculus]] |
| − | [[ | + | [[Category:Integration]] |
Latest revision as of 21:17, September 8, 2020
|
This article/section deals with mathematical concepts appropriate for late high school or early college. |
This article details the method known as Integration by Parts.
Contents
Integration by Parts
Integration by parts is a special technique to facilitate the integration of the product of two functions that otherwise lack an obvious integral. This technique can be proven with the product rule.
The rule for integration by parts is stated as follows:[1]
This rule is often useful when one function is a power of x and the other function is a trigonometric function or e raised to a power of x.
Note that it may be necessary to repeat the integration by parts several times, depending on the situation.
Proving Integration by Parts
The proof for Integration by Parts is simple and important. Given:
It is known from the Product rule that:
If both sides are integrated:
By rearranging the terms:
Sometimes, Integration by Parts is simply expressed as:
Using Integration by Parts
Example 1
A common example where integration by parts is needed is this:
Let's set:
We can begin to solve:
Now we are left with an integral that we do know how to solve:
Therefore:
Example 2
Say that we are given:
Let's set:
And continue solving:
And continue solving:
Now, we set:
And continue solving:
Let's look back at what we have done:
The integral of the function has the integral of itself ... inside itself. This is actually a common occurrence with functions like the one we are solving. However, we can cheat the never-ending cycle of integrals by doing this:
Rapid Repeated Integration
Rapid Repeated Integration is a shortcut method for reduction problems that require Integration by Parts. It is especially useful when one function's derivative reduces to zero.
Example 3
For example, integrating:
We start off by making a table of 
and 
. On the first column, the derivatives of are taken until they reach zero. In the second column, sin(x) is integrated once down each row:
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Terms are multiplied diagonally from the left to the right and we add the product to the product of the next product, alternating signs with each step.
The first term, for example, is:
So then the integral becomes:
= 
- =
While this method is very useful when part of the integrand becomes zero through derivation, it does not work in cases such as Example 2 above.
