MAT 1235
Calculus II
Section 7.1
Integration By Parts
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Exam II
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Exam II
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Preview
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Finding antiderivatives are way-way-way
more difficult than finding derivatives.
We need to develop more techniques to
help finding antiderivatives.
IBP is considered as the reverse process
of the product rule.
Product Rule
d
 f ( x) g ( x)  f ( x) g ( x)  f ( x) g ( x)
dx
  f ( x) g ( x)  f ( x) g ( x)dx  f ( x) g ( x)
 f ( x) g ( x)dx   f ( x) g ( x)dx  f ( x) g ( x)
 f ( x) g ( x)dx  f ( x) g ( x)   f ( x) g ( x)dx
Product Rule
d
 f ( x) g ( x)  f ( x) g ( x)  f ( x) g ( x)
dx
  f ( x) g ( x)  f ( x) g ( x)dx  f ( x) g ( x)
 f ( x) g ( x)dx   f ( x) g ( x)dx  f ( x) g ( x)
 f ( x) g ( x)dx  f ( x) g ( x)   f ( x) g ( x)dx
Definition of Antiderivatives
d
 f ( x) g ( x)  f ( x) g ( x)  f ( x) g ( x)
dx
  f ( x) g ( x)  f ( x) g ( x)dx  f ( x) g ( x)
 f ( x) g ( x)dx   f ( x) g ( x)dx  f ( x) g ( x)
 f ( x) g ( x)dx  f ( x) g ( x)   f ( x) g ( x)dx
Rationale
Easier
Difficult

f ( x) g ( x)dx  f ( x) g ( x)   f ( x) g ( x)dx
The integral on the right hand side is
easier to evaluate than the one on the left
hand side
Rationale
Easier
Difficult

f ( x) g ( x)dx  f ( x) g ( x)   f ( x) g ( x)dx
 x cos xdx  x sin x   sin xdx
The integral on the right hand side is
easier to evaluate than the one on the left
hand side
Alternative Form
 f ( x) g ( x)dx  f ( x) g ( x)   f ( x) g ( x)dx
u  f ( x), v  g ( x)
Then du  f ( x)dx, dv  g ( x)dx
Let
We have
 udv  uv   vdu
Integration By Parts
f ( x) g ( x)dx  f ( x) g ( x)   f ( x) g ( x)dx

 udv  uv   vdu
b
b
 f ( x) g ( x)dx   f ( x) g ( x)   f ( x) g ( x)dx
b
a
a
a
Example 1
 x cos xdx
Example 1(Analysis)
 udv  uv   vdu
 x cos xdx
In order to use IBP, we need to choose
𝑢 and 𝑑𝑣
One possible choice is u  x, dv  cos xdx
Example 1(Analysis)
 x cos xdx
In order to use IBP, we need to choose
𝑢 and 𝑑𝑣
 udv  uv   vdu
One possible choice is
Let u  x, dv  cos xdx
Example 1(Analysis)
 x cos xdx
Let
u  x, dv  cos xdx
Then, du  dx, v   cos xdx
 sin x
Now, we need to find 𝑣 and 𝑑𝑢
 udv  uv   vdu
Example 1(Analysis)
 x cos xdx
Let
u  x, dv  cos xdx
Then, du  dx, v   cos xdx
 sin x
Now, we need to find 𝑣 and 𝑑𝑢
 udv  uv   vdu
Example 1
 x cos xdx
 x sin x   sin xdx

Let
u  x, dv  cos xdx
Then, du  dx, v   cos xdx
 sin x
 udv  uv   vdu
 udv  uv   vdu
Example 1
 x cos xdx
 x sin x   sin xdx

Let
u  x, dv  cos xdx
Then, du  dx, v   cos xdx
 sin x  B
What if?
 udv  uv   vdu
Example 1
 x cos xdx
   cos x  xdx

Let
u  cos x, dv  xdx
x2
Then, du   sin xdx, v   xdx 
2
What if?
Example 2
x
xe
 dx
x
xe
 dx 
 udv  uv   vdu
Expectations
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All supporting steps are required and
done on the right side.
Example 3
2
 ln xdx
1
2
 ln xdx 
1
 udv  uv   vdu
Example 4 (Classical Example)
x
e
 cos xdx
x
e
 cos xdx 
ex
 udv  uv   vdu