text: Integration By Parts
  
  Contributed by:
  This presentation discusses integration by parts with some relevant examples.
  1. 6.3 Integration By Parts
  Badlands, South Dakota
  Photo by Vickie Kelly, 1993 Greg Kelly, Hanford High School, Richland, Washington
  2. 6.3 Integration By Parts
  Start with the product rule:
  u dv  d  uv   v du 
  d dv du
   uv  u  v
  dx dx dx
  u dv  d  uv    v du
  d  uv  u dv  v du
  d  uv   v du u dv u dv uv  v du
  This is the Integration by Parts
  u dv d  uv   v du formula.
  
  3. u dv uv  v du
  dv is easy to
  u differentiates to integrate.
  zero (usually).
  The Integration by Parts formula is a “product rule” for
  integration.
  Choose u in this order: LIPET
  Logs, Inverse trig, Polynomial, Exponential, Trig
  
  4. Example 1:
  u dv uv  v du
  x cos x dx LIPET
  polynomial factor u x dv cos x dx
  du dx v sin x
  u v  v du
  x sin x  sin x dx
  x sin x  cos x  C
  
  5. u dv uv  v du
  ln x dx LIPET
  logarithmic factor u ln x dv dx
  u v  v du 1
  du  dx v x
  x
  1
  ln x x  x  dx
  x
  x ln x  x  C
  
  6. Example 4:
  2 x u dv uv  v du LIPET
  x e dx
  u x 2 dv e x dx
  u v  v du
  du 2 x dx v e x
  2 x x
  x e  e 2 x dx
  2 x x This is still a product, so we
  x e  2xe dx u
  need to use x integration
  dv e dxx
  by
  parts again.
  du dx v e x
  2 x
  
  x e  2 xe  e dx x x
  
  2 x x x
  x e  2 xe  2e  C
  
  7. Example 5: LIPET
  x u e x dv cos x dx
   cos x dx
  e
  du e x dx v sin x
  u v  v du x
  u e dv sin x dx
  x x
  e sin x  sin x e dx x
  du e dx v  cos x
  x
   x
  e sin x  e  cos x   cos x e dx x
  
  uv v du This is the
  x x x
  e sin x  e cos x  e cos x dx expression we
  started with!
  
  8. Example 6: LIPET
  x u e x dv cos x dx
   cos x dx
  e
  du e x dx v sin x
  u v  v du x
  u e dv sin x dx
  x x
  e sin x  sin x e dx x
  du e dx v  cos x
  x
   x
  e sin x  e  cos x   cos x e dx x
  
  x x x x
  e cos x dx e sin x  e cos x  e cos x dx
  x x x
  2 e cos x dx e sin x  e cos x
  x x
  x e sin x  e cos x
  e cos x dx  2
  C
  9. Example 6:
  x
  This is called “solving for
  e cos x dx the unknown integral.”
  u v  v du It works when both
  x x factors integrate and
  e sin x  sin x e dx differentiate forever.
  x
   x
  e sin x  e  cos x   cos x e dx x
  
  x x x x
  e cos x dx e sin x  e cos x  e cos x dx
  x x x
  2 e cos x dx e sin x  e cos x
  x x
  x e sin x  e cos x
  e cos x dx  2
  C
  
  10. A Shortcut: Tabular Integration
  Tabular integration works for integrals of the form:
  f  x  g  x  dx
  where: Differentiates to Integrates
  zero in several repeatedly.
  steps.
  
  11. 2 x
   e dx
  x
  f  x  & deriv. g  x  & integrals
  2 x
   x e
   2x ex
  Compare this with
   2 e x the same problem
  done the other way:
  x
  0 e
  2 x x x
  2 x
  x e dx  x e  2 xe 2e C
  
  12. Example 5:
  2 x u dv uv  v du LIPET
  x e dx
  u x 2 dv e x dx
  u v  v du
  du 2 x dx v e x
  2 x x
  x e  e 2 x dx
  x 2 e x  2xe x dx u x dv e dxx
  du dx v e x
  2 x
  
  x e  2 xe  e dx x x
  
  This is easier and quicker to
  2 x x x
  x e  2 xe  2e  C do with tabular integration!
  
  13. 3
   sin x dx
  x
  3
   x sin x
   3x 2  cos x
   6x  sin x
   6 cos x
  0 sin x
   x 3 cos x  3 x 2 sin x  6 x cos x  6sin x + C
explanation: The document is a presentation on integration by parts, which includes several examples and explanations of the technique. It does not contain a single question or a question and answer; instead, it provides a detailed explanation and multiple examples of how to use integration by parts.
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distilabel_metadata.content: Integration By Parts
  
  Contributed by:
  This presentation discusses integration by parts with some relevant examples.
  1. 6.3 Integration By Parts
  Badlands, South Dakota
  Photo by Vickie Kelly, 1993 Greg Kelly, Hanford High School, Richland, Washington
  2. 6.3 Integration By Parts
  Start with the product rule:
  u dv  d  uv   v du 
  d dv du
   uv  u  v
  dx dx dx
  u dv  d  uv    v du
  d  uv  u dv  v du
  d  uv   v du u dv u dv uv  v du
  This is the Integration by Parts
  u dv d  uv   v du formula.
  
  3. u dv uv  v du
  dv is easy to
  u differentiates to integrate.
  zero (usually).
  The Integration by Parts formula is a “product rule” for
  integration.
  Choose u in this order: LIPET
  Logs, Inverse trig, Polynomial, Exponential, Trig
  
  4. Example 1:
  u dv uv  v du
  x cos x dx LIPET
  polynomial factor u x dv cos x dx
  du dx v sin x
  u v  v du
  x sin x  sin x dx
  x sin x  cos x  C
  
  5. u dv uv  v du
  ln x dx LIPET
  logarithmic factor u ln x dv dx
  u v  v du 1
  du  dx v x
  x
  1
  ln x x  x  dx
  x
  x ln x  x  C
  
  6. Example 4:
  2 x u dv uv  v du LIPET
  x e dx
  u x 2 dv e x dx
  u v  v du
  du 2 x dx v e x
  2 x x
  x e  e 2 x dx
  2 x x This is still a product, so we
  x e  2xe dx u
  need to use x integration
  dv e dxx
  by
  parts again.
  du dx v e x
  2 x
  
  x e  2 xe  e dx x x
  
  2 x x x
  x e  2 xe  2e  C
  
  7. Example 5: LIPET
  x u e x dv cos x dx
   cos x dx
  e
  du e x dx v sin x
  u v  v du x
  u e dv sin x dx
  x x
  e sin x  sin x e dx x
  du e dx v  cos x
  x
   x
  e sin x  e  cos x   cos x e dx x
  
  uv v du This is the
  x x x
  e sin x  e cos x  e cos x dx expression we
  started with!
  
  8. Example 6: LIPET
  x u e x dv cos x dx
   cos x dx
  e
  du e x dx v sin x
  u v  v du x
  u e dv sin x dx
  x x
  e sin x  sin x e dx x
  du e dx v  cos x
  x
   x
  e sin x  e  cos x   cos x e dx x
  
  x x x x
  e cos x dx e sin x  e cos x  e cos x dx
  x x x
  2 e cos x dx e sin x  e cos x
  x x
  x e sin x  e cos x
  e cos x dx  2
  C
  9. Example 6:
  x
  This is called “solving for
  e cos x dx the unknown integral.”
  u v  v du It works when both
  x x factors integrate and
  e sin x  sin x e dx differentiate forever.
  x
   x
  e sin x  e  cos x   cos x e dx x
  
  x x x x
  e cos x dx e sin x  e cos x  e cos x dx
  x x x
  2 e cos x dx e sin x  e cos x
  x x
  x e sin x  e cos x
  e cos x dx  2
  C
  
  10. A Shortcut: Tabular Integration
  Tabular integration works for integrals of the form:
  f  x  g  x  dx
  where: Differentiates to Integrates
  zero in several repeatedly.
  steps.
  
  11. 2 x
   e dx
  x
  f  x  & deriv. g  x  & integrals
  2 x
   x e
   2x ex
  Compare this with
   2 e x the same problem
  done the other way:
  x
  0 e
  2 x x x
  2 x
  x e dx  x e  2 xe 2e C
  
  12. Example 5:
  2 x u dv uv  v du LIPET
  x e dx
  u x 2 dv e x dx
  u v  v du
  du 2 x dx v e x
  2 x x
  x e  e 2 x dx
  x 2 e x  2xe x dx u x dv e dxx
  du dx v e x
  2 x
  
  x e  2 xe  e dx x x
  
  This is easier and quicker to
  2 x x x
  x e  2 xe  2e  C do with tabular integration!
  
  13. 3
   sin x dx
  x
  3
   x sin x
   3x 2  cos x
   6x  sin x
   6 cos x
  0 sin x
   x 3 cos x  3 x 2 sin x  6 x cos x  6sin x + C
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document is a presentation on integration by parts, which includes several examples and explanations of the technique. It does not contain a single question or a question and answer; instead, it provides a detailed explanation and multiple examples of how to use integration by parts.
  
  Label: no-question