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Integration by parts (view source)
Revision as of 17:41, December 31, 2008
, 17:41, December 31, 2008added another example and fixed minor errors
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, one for each 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===
===Proving Integration by Parts===
<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 ===
This is a relatively strange case that comes up often.
<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==