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Integration by Parts


Integration by Parts is a special method of integration that is often useful when two functions are multiplied together, but is also ...

Integration by Parts is a special method of integration that is often useful when two functions are multiplied together, but is also helpful in other ways.
You will see plenty of examples soon, but first let us see the rule:
u v dx = uv dx −u' (v dx) dx
  • u is the function u(x)
  • v is the function v(x)
As a diagram:
integration by parts general
Let's get straight into an example, and talk about it after:

Example: What is x cos(x) dx ?

OK, we have x multiplied by cos(x), so integration by parts is a good choice.
First choose which functions for u and v:
  • u = x
  • v = cos(x)
So now it is in the format u v dx we can proceed:
Differentiate uu' = x' = 1
Integrate vv dx = cos(x) dx = sin(x)   (see Integration Rules)
Now we can put it together:
integration by parts x cos(x) dx
Simplify and solve:
x sin(x) − sin(x) dx
x sin(x) + cos(x) + C

So we followed these steps:
  • Choose u and v
  • Differentiate u: u'
  • Integrate v: v dx
  • Put u, u' and ∫v dx into: uv dx −u' (v dx) dx
  • Simplify and solve
In English, to help you remember, u v dx becomes:
(u integral v) minus integral of (derivative u, integral v)

Let's try some more examples:

Example: What is ∫ln(x)/x2 dx ?

First choose u and v:
  • u = ln(x)
  • v = 1/x2
Differentiate u: ln(x)' = 1/x
Integrate v: 1/x2 dx = x-2 dx = −x-1 = -1/x   (by the power rule)
Now put it together:
integration by parts ln(x) on x^2
−ln(x)/x − ∫1/x2 dx = −ln(x)/x − 1/x + C
−(ln(x) + 1)/x + C

Example: What is ln(x) dx ?

But there is only one function! How do we choose u and v ?
Hey! We can just choose v as being "1":
  • u = ln(x)
  • v = 1
Differentiate u: ln(x)' = 1/x
Integrate v: 1 dx = x
Now put it together:
integration by parts ln(x)
x ln(x) − 1 dx = x ln(x) − x + C

Example: What is ∫ex x dx ?

Choose u and v:
  • u = ex
  • v = x
Differentiate u: (ex)' = ex
Integrate v: x dx = x2/2
Now put it together:
integration by parts e^x x
Well, that was a spectacular disaster! It just got more complicated.
Maybe we could choose a different u and v?

Example: ex x dx (continued)

Choose u and v differently:
  • u = x
  • v = ex
Differentiate u: (x)' = 1
Integrate v: ex dx = ex
Now put it together:
integration by parts x e^x
ex − ex + C
ex(x−1) + C
The moral of the story: Choose u and v carefully!
Choose a u that gets simpler when you differentiate it and a v that doesn't get any more complicated when you integrate it.
A helpful rule of thumb is I LATE. Choose u based on which of these comes first:
  • I: Inverse trigonometric functions such as sin-1(x), cos-1(x), tan-1(x)
  • L: Logarithmic functions such as ln(x), log(x)
  • A: Algebraic functions such as x2, x3
  • T: Trigonometric functions such as sin(x), cos(x), tan (x)
  • E: Exponential functions such as ex, 3x

And here is one last (and tricky) example:

Example: ex sin(x) dx

Choose u and v:
  • u = sin(x)
  • v = ex
Differentiate u: sin(x)' = cos(x)
Integrate v: ex dx = ex
Now put it together:
ex sin(x) dx = sin(x) ex -cos(x) ex dx

Looks worse, but let us persist! We can use integration by parts again:
Choose u and v:
  • u = cos(x)
  • v = ex
Differentiate u: cos(x)' = -sin(x)
Integrate v: ex dx = ex
Now put it together:
ex sin(x) dx = sin(x) ex - (cos(x) ex −−sin(x) ex dx)
ex sin(x) dx = ex sin(x) - ex cos(x) − ex sin(x)dx
Now we have the same integral on both sides (except one is subtracted) ...
... so bring the right hand one over to the left and we get:
2ex sin(x) dx = ex sin(x) − ex cos(x)
ex sin(x) dx = ex (sin(x) - cos(x)) / 2 + C

Footnote: Where Did "Integration by Parts" Come From?

It is based on the Product Rule for Derivatives:
(uv)' = uv' + u'v
Integrate both sides and rearrange:
(uv)' dx = uv' dx + u'v dx
uv = uv' dx + u'v dx
uv' dx = uv − u'v dx
Some people prefer that last form, but I like to integrate v' so the left side is simple:
uv dx = uv dx − u'(v dx) dx



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Ilm Gah: Integration by Parts
Integration by Parts
Ilm Gah
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