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Integration by Parts, Part 3

The document discusses the LIATE rule for integration by parts. LIATE is an acronym that indicates the order of priority for choosing the u(x) function: logarithmic, inverse trigonometric, algebraic, trigonometric, exponential. Two examples are provided to illustrate applying the LIATE rule. In the first example, choosing u(x)=x for x sin x dx works according to LIATE. In the second example, choosing u(x)=sin x for ex sin x dx works according to LIATE after solving using integration by parts twice.

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The LIATE Rule
The LIATE Rule The difficulty of integration by parts is in choosing u(x) and v (x) correctly.
The LIATE Rule The difficulty of integration by parts is in choosing u(x) and v (x) correctly. The LIATE rule is a rule of thumb that tells you which function you should choose as u(x):
The LIATE Rule The difficulty of integration by parts is in choosing u(x) and v (x) correctly. The LIATE rule is a rule of thumb that tells you which function you should choose as u(x): LIATE
The LIATE Rule The difficulty of integration by parts is in choosing u(x) and v (x) correctly. The LIATE rule is a rule of thumb that tells you which function you should choose as u(x): L Log IATE
The LIATE Rule The difficulty of integration by parts is in choosing u(x) and v (x) correctly. The LIATE rule is a rule of thumb that tells you which function you should choose as u(x): L I Inv. Trig ATE
The LIATE Rule The difficulty of integration by parts is in choosing u(x) and v (x) correctly. The LIATE rule is a rule of thumb that tells you which function you should choose as u(x): LI A Algebraic TE
The LIATE Rule The difficulty of integration by parts is in choosing u(x) and v (x) correctly. The LIATE rule is a rule of thumb that tells you which function you should choose as u(x): LIA T Trig E
The LIATE Rule The difficulty of integration by parts is in choosing u(x) and v (x) correctly. The LIATE rule is a rule of thumb that tells you which function you should choose as u(x): LIAT E Exp
The LIATE Rule The difficulty of integration by parts is in choosing u(x) and v (x) correctly. The LIATE rule is a rule of thumb that tells you which function you should choose as u(x): LIATE
The LIATE Rule The difficulty of integration by parts is in choosing u(x) and v (x) correctly. The LIATE rule is a rule of thumb that tells you which function you should choose as u(x): LIATE The word itself tells you in which order of priority you should use u(x).
Example 1
Example 1 x sin xdx
Example 1 x sin xdx Here we have an algebraic and a trigonometric function:
Example 1 x sin xdx Here we have an algebraic and a trigonometric function: x sin xdx
Example 1 x sin xdx Here we have an algebraic and a trigonometric function:   A x sin xdx
Example 1 x sin xdx Here we have an algebraic and a trigonometric function: x$$$XT sin xdx
Example 1 x sin xdx Here we have an algebraic and a trigonometric function: x sin xdx
Example 1 x sin xdx Here we have an algebraic and a trigonometric function: x sin xdx According to LIATE, the A comes before the T, so we choose u(x) = x.
Example 2
Example 2 ex sin xdx
Example 2 ex sin xdx In this case we have E and T:
Example 2 ex sin xdx In this case we have E and T: b E ex sin xdx
Example 2 ex sin xdx In this case we have E and T: ex$$$XT sin xdx
Example 2 ex sin xdx In this case we have E and T: ex sin xdx
Example 2 ex sin xdx In this case we have E and T: ex sin xdx According to LIATE, the T comes before the E, so we choose u(x) = sin x.
Example 2 ex sin xdx In this case we have E and T: ex sin xdx According to LIATE, the T comes before the E, so we choose u(x) = sin x. Let’s solve this integral to show that this works:
Example 2 ex sin xdx In this case we have E and T: ex sin xdx According to LIATE, the T comes before the E, so we choose u(x) = sin x. Let’s solve this integral to show that this works: u(x) = sin x
Example 2 ex sin xdx In this case we have E and T: ex sin xdx According to LIATE, the T comes before the E, so we choose u(x) = sin x. Let’s solve this integral to show that this works: u(x) = sin x v (x) = ex
Example 2 ex sin xdx In this case we have E and T: ex sin xdx According to LIATE, the T comes before the E, so we choose u(x) = sin x. Let’s solve this integral to show that this works: u(x) = sin x v (x) = ex u (x) = cos x
Example 2 ex sin xdx In this case we have E and T: ex sin xdx According to LIATE, the T comes before the E, so we choose u(x) = sin x. Let’s solve this integral to show that this works: u(x) = sin x v (x) = ex u (x) = cos x v(x) = ex
Example 2 ex sin xdx In this case we have E and T: ex sin xdx According to LIATE, the T comes before the E, so we choose u(x) = sin x. Let’s solve this integral to show that this works: u(x) = sin x v (x) = ex u (x) = cos x v(x) = ex ex sin xdx = ex sin x − ex cos xdx
Example 2 ex sin xdx In this case we have E and T: ex sin xdx According to LIATE, the T comes before the E, so we choose u(x) = sin x. Let’s solve this integral to show that this works: u(x) = sin x v (x) = ex u (x) = cos x v(x) = ex b v (x) ex sin xdx = ex sin x − ex cos xdx
Example 2 ex sin xdx In this case we have E and T: ex sin xdx According to LIATE, the T comes before the E, so we choose u(x) = sin x. Let’s solve this integral to show that this works: u(x) = sin x v (x) = ex u (x) = cos x v(x) = ex ex$$$Xu(x) sin xdx = ex sin x − ex cos xdx
Example 2 ex sin xdx In this case we have E and T: ex sin xdx According to LIATE, the T comes before the E, so we choose u(x) = sin x. Let’s solve this integral to show that this works: u(x) = sin x v (x) = ex u (x) = cos x v(x) = ex ex sin xdx = b v(x) ex sin x − ex cos xdx
Example 2 ex sin xdx In this case we have E and T: ex sin xdx According to LIATE, the T comes before the E, so we choose u(x) = sin x. Let’s solve this integral to show that this works: u(x) = sin x v (x) = ex u (x) = cos x v(x) = ex ex sin xdx = ex$$$Xu(x) sin x − ex cos xdx
Example 2 ex sin xdx In this case we have E and T: ex sin xdx According to LIATE, the T comes before the E, so we choose u(x) = sin x. Let’s solve this integral to show that this works: u(x) = sin x v (x) = ex u (x) = cos x v(x) = ex ex sin xdx = ex sin x − b v(x) ex cos xdx
Example 2 ex sin xdx In this case we have E and T: ex sin xdx According to LIATE, the T comes before the E, so we choose u(x) = sin x. Let’s solve this integral to show that this works: u(x) = sin x v (x) = ex u (x) = cos x v(x) = ex ex sin xdx = ex sin x − ex $$$Xu (x) cos xdx
Example 2 ex sin xdx In this case we have E and T: ex sin xdx According to LIATE, the T comes before the E, so we choose u(x) = sin x. Let’s solve this integral to show that this works: u(x) = sin x v (x) = ex u (x) = cos x v(x) = ex ex sin xdx = ex sin x − ex cos xdx
Example 2
Example 2 Now we need to solve that integral:
Example 2 Now we need to solve that integral: ex sin xdx = ex sin x − ex cos xdx
Example 2 Now we need to solve that integral: ex sin xdx = ex sin x − ex cos xdx We need to use integration by parts again.
Example 2 Now we need to solve that integral: ex sin xdx = ex sin x − ex cos xdx We need to use integration by parts again. By using LIATE, we choose u(x) = cos x:
Example 2 Now we need to solve that integral: ex sin xdx = ex sin x − ex cos xdx We need to use integration by parts again. By using LIATE, we choose u(x) = cos x: u(x) = cos x
Example 2 Now we need to solve that integral: ex sin xdx = ex sin x − ex cos xdx We need to use integration by parts again. By using LIATE, we choose u(x) = cos x: u(x) = cos x v (x) = ex
Example 2 Now we need to solve that integral: ex sin xdx = ex sin x − ex cos xdx We need to use integration by parts again. By using LIATE, we choose u(x) = cos x: u(x) = cos x v (x) = ex u (x) = − sin x
Example 2 Now we need to solve that integral: ex sin xdx = ex sin x − ex cos xdx We need to use integration by parts again. By using LIATE, we choose u(x) = cos x: u(x) = cos x v (x) = ex u (x) = − sin x v(x) = ex
Example 2 Now we need to solve that integral: ex sin xdx = ex sin x − ex cos xdx We need to use integration by parts again. By using LIATE, we choose u(x) = cos x: u(x) = cos x v (x) = ex u (x) = − sin x v(x) = ex ex sin xdx = ex sin x − ex cos xdx
Example 2 Now we need to solve that integral: ex sin xdx = ex sin x − ex cos xdx We need to use integration by parts again. By using LIATE, we choose u(x) = cos x: u(x) = cos x v (x) = ex u (x) = − sin x v(x) = ex ex sin xdx = ex sin x − b v (x) ex cos xdx
Example 2 Now we need to solve that integral: ex sin xdx = ex sin x − ex cos xdx We need to use integration by parts again. By using LIATE, we choose u(x) = cos x: u(x) = cos x v (x) = ex u (x) = − sin x v(x) = ex ex sin xdx = ex sin x − ex $$$Xu(x) cos xdx
Example 2 Now we need to solve that integral: ex sin xdx = ex sin x − ex cos xdx We need to use integration by parts again. By using LIATE, we choose u(x) = cos x: u(x) = cos x v (x) = ex u (x) = − sin x v(x) = ex ex sin xdx = ex sin x − ex cos xdx
Example 2 Now we need to solve that integral: ex sin xdx = ex sin x − ex cos xdx We need to use integration by parts again. By using LIATE, we choose u(x) = cos x: u(x) = cos x v (x) = ex u (x) = − sin x v(x) = ex ex sin xdx = ex sin x − ex cos xdx = ex sin x − ex cos x − ex (− sin x) dx
Example 2 Now we need to solve that integral: ex sin xdx = ex sin x − ex cos xdx We need to use integration by parts again. By using LIATE, we choose u(x) = cos x: u(x) = cos x v (x) = ex u (x) = − sin x v(x) = ex ex sin xdx = ex sin x − ex cos xdx = ex sin x − b v(x) ex cos x − ex (− sin x) dx
Example 2 Now we need to solve that integral: ex sin xdx = ex sin x − ex cos xdx We need to use integration by parts again. By using LIATE, we choose u(x) = cos x: u(x) = cos x v (x) = ex u (x) = − sin x v(x) = ex ex sin xdx = ex sin x − ex cos xdx = ex sin x − ex $$$Xu(x) cos x − ex (− sin x) dx
Example 2 Now we need to solve that integral: ex sin xdx = ex sin x − ex cos xdx We need to use integration by parts again. By using LIATE, we choose u(x) = cos x: u(x) = cos x v (x) = ex u (x) = − sin x v(x) = ex ex sin xdx = ex sin x − ex cos xdx = ex sin x − ex cos x − b v(x) ex (− sin x) dx
Example 2 Now we need to solve that integral: ex sin xdx = ex sin x − ex cos xdx We need to use integration by parts again. By using LIATE, we choose u(x) = cos x: u(x) = cos x v (x) = ex u (x) = − sin x v(x) = ex ex sin xdx = ex sin x − ex cos xdx = ex sin x − ex cos x − ex $$$$$Xu (x) (− sin x)dx
Example 2 Now we need to solve that integral: ex sin xdx = ex sin x − ex cos xdx We need to use integration by parts again. By using LIATE, we choose u(x) = cos x: u(x) = cos x v (x) = ex u (x) = − sin x v(x) = ex ex sin xdx = ex sin x − ex cos xdx = ex sin x − ex cos x − ex (− sin x) dx
Example 2
Example 2 So, we have that:
Example 2 So, we have that: ex sin xdx = ex sin x − ex cos x − ex sin xdx
Example 2 So, we have that: ex sin xdx !! = ex sin x − ex cos x − ex sin xdx !!
Example 2 So, we have that: ex sin xdx !! = ex sin x − ex cos x − ex sin xdx !! So, we add ex sin xdx to both sides of the equation:
Example 2 So, we have that: ex sin xdx !! = ex sin x − ex cos x − ex sin xdx !! So, we add ex sin xdx to both sides of the equation: 2 ex sin xdx = ex (sin x − cos x)
Example 2 So, we have that: ex sin xdx !! = ex sin x − ex cos x − ex sin xdx !! So, we add ex sin xdx to both sides of the equation: 2 ex sin xdx = ex (sin x − cos x) ex sin xdx = ex 2 (sin x − cos x) + C
Example 2 So, we have that: ex sin xdx !! = ex sin x − ex cos x − ex sin xdx !! So, we add ex sin xdx to both sides of the equation: 2 ex sin xdx = ex (sin x − cos x) ex sin xdx = ex 2 (sin x − cos x) + C This is in fact an example of Trick number 2.
Example 3
Example 3 Let’s do another example:
Example 3 Let’s do another example: x2 ex dx
Example 3 Let’s do another example: x2 ex dx We have an A and an E:
Example 3 Let’s do another example: x2 ex dx We have an A and an E:    A x2 ex dx
Example 3 Let’s do another example: x2 ex dx We have an A and an E: x2 b E ex dx
Example 3 Let’s do another example: x2 ex dx We have an A and an E: x2 ex dx
Example 3 Let’s do another example: x2 ex dx We have an A and an E: x2 ex dx LIATE says that the A comes before the E. So, we choose u(x) = x2:
Example 3 Let’s do another example: x2 ex dx We have an A and an E: x2 ex dx LIATE says that the A comes before the E. So, we choose u(x) = x2: u(x) = x2
Example 3 Let’s do another example: x2 ex dx We have an A and an E: x2 ex dx LIATE says that the A comes before the E. So, we choose u(x) = x2: u(x) = x2 v (x) = ex
Example 3 Let’s do another example: x2 ex dx We have an A and an E: x2 ex dx LIATE says that the A comes before the E. So, we choose u(x) = x2: u(x) = x2 v (x) = ex u (x) = 2x
Example 3 Let’s do another example: x2 ex dx We have an A and an E: x2 ex dx LIATE says that the A comes before the E. So, we choose u(x) = x2: u(x) = x2 v (x) = ex u (x) = 2x v(x) = ex
Example 3 Let’s do another example: x2 ex dx We have an A and an E: x2 ex dx LIATE says that the A comes before the E. So, we choose u(x) = x2: u(x) = x2 v (x) = ex u (x) = 2x v(x) = ex x2 ex dx =
Example 3 Let’s do another example: x2 ex dx We have an A and an E: x2 ex dx LIATE says that the A comes before the E. So, we choose u(x) = x2: u(x) = x2 v (x) = ex u (x) = 2x v(x) = ex    u(x) x2 ex dx =
Example 3 Let’s do another example: x2 ex dx We have an A and an E: x2 ex dx LIATE says that the A comes before the E. So, we choose u(x) = x2: u(x) = x2 v (x) = ex u (x) = 2x v(x) = ex x2 b v (x) ex dx =
Example 3 Let’s do another example: x2 ex dx We have an A and an E: x2 ex dx LIATE says that the A comes before the E. So, we choose u(x) = x2: u(x) = x2 v (x) = ex u (x) = 2x v(x) = ex x2 ex dx = x2 ex − 2xex dx
Example 3 Let’s do another example: x2 ex dx We have an A and an E: x2 ex dx LIATE says that the A comes before the E. So, we choose u(x) = x2: u(x) = x2 v (x) = ex u (x) = 2x v(x) = ex x2 ex dx =    u(x) x2 ex − 2xex dx
Example 3 Let’s do another example: x2 ex dx We have an A and an E: x2 ex dx LIATE says that the A comes before the E. So, we choose u(x) = x2: u(x) = x2 v (x) = ex u (x) = 2x v(x) = ex x2 ex dx = x2 b v(x) ex − 2xex dx
Example 3 Let’s do another example: x2 ex dx We have an A and an E: x2 ex dx LIATE says that the A comes before the E. So, we choose u(x) = x2: u(x) = x2 v (x) = ex u (x) = 2x v(x) = ex x2 ex dx = x2 ex − b u (x) 2xex dx
Example 3 Let’s do another example: x2 ex dx We have an A and an E: x2 ex dx LIATE says that the A comes before the E. So, we choose u(x) = x2: u(x) = x2 v (x) = ex u (x) = 2x v(x) = ex x2 ex dx = x2 ex − 2xb v(x) ex dx
Example 3 Let’s do another example: x2 ex dx We have an A and an E: x2 ex dx LIATE says that the A comes before the E. So, we choose u(x) = x2: u(x) = x2 v (x) = ex u (x) = 2x v(x) = ex x2 ex dx = x2 ex − 2xex dx
Example 3
Example 3 x2 ex dx = x2 ex − 2 xex dx
Example 3 x2 ex dx = x2 ex − 2 xex dx Now we need to solve that integral by integration by parts again:
Example 3 x2 ex dx = x2 ex − 2 xex dx Now we need to solve that integral by integration by parts again: u(x) = x
Example 3 x2 ex dx = x2 ex − 2 xex dx Now we need to solve that integral by integration by parts again: u(x) = x v (x) = ex
Example 3 x2 ex dx = x2 ex − 2 xex dx Now we need to solve that integral by integration by parts again: u(x) = x v (x) = ex u (x) = 1
Example 3 x2 ex dx = x2 ex − 2 xex dx Now we need to solve that integral by integration by parts again: u(x) = x v (x) = ex u (x) = 1 v(x) = ex
Example 3 x2 ex dx = x2 ex − 2 xex dx Now we need to solve that integral by integration by parts again: u(x) = x v (x) = ex u (x) = 1 v(x) = ex x2 ex dx = x2 ex − 2 xex − 1.ex dx
Example 3 x2 ex dx = x2 ex − 2 xex dx Now we need to solve that integral by integration by parts again: u(x) = x v (x) = ex u (x) = 1 v(x) = ex x2 ex dx = x2 ex − 2 xex − 1.ex dx = x2 ex − 2xex + 2ex = ex x2 − 2x + 2
Example 3 x2 ex dx = x2 ex − 2 xex dx Now we need to solve that integral by integration by parts again: u(x) = x v (x) = ex u (x) = 1 v(x) = ex x2 ex dx = x2 ex − 2 xex − 1.ex dx = x2 ex − 2xex + 2ex = ex x2 − 2x + 2
Integration by Parts, Part 3

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Integration by Parts, Part 3

  • 3.
  • 4.
    The LIATE Rule Thedifficulty of integration by parts is in choosing u(x) and v (x) correctly.
  • 5.
    The LIATE Rule Thedifficulty of integration by parts is in choosing u(x) and v (x) correctly. The LIATE rule is a rule of thumb that tells you which function you should choose as u(x):
  • 6.
    The LIATE Rule Thedifficulty of integration by parts is in choosing u(x) and v (x) correctly. The LIATE rule is a rule of thumb that tells you which function you should choose as u(x): LIATE
  • 7.
    The LIATE Rule Thedifficulty of integration by parts is in choosing u(x) and v (x) correctly. The LIATE rule is a rule of thumb that tells you which function you should choose as u(x): L Log IATE
  • 8.
    The LIATE Rule Thedifficulty of integration by parts is in choosing u(x) and v (x) correctly. The LIATE rule is a rule of thumb that tells you which function you should choose as u(x): L I Inv. Trig ATE
  • 9.
    The LIATE Rule Thedifficulty of integration by parts is in choosing u(x) and v (x) correctly. The LIATE rule is a rule of thumb that tells you which function you should choose as u(x): LI A Algebraic TE
  • 10.
    The LIATE Rule Thedifficulty of integration by parts is in choosing u(x) and v (x) correctly. The LIATE rule is a rule of thumb that tells you which function you should choose as u(x): LIA T Trig E
  • 11.
    The LIATE Rule Thedifficulty of integration by parts is in choosing u(x) and v (x) correctly. The LIATE rule is a rule of thumb that tells you which function you should choose as u(x): LIAT E Exp
  • 12.
    The LIATE Rule Thedifficulty of integration by parts is in choosing u(x) and v (x) correctly. The LIATE rule is a rule of thumb that tells you which function you should choose as u(x): LIATE
  • 13.
    The LIATE Rule Thedifficulty of integration by parts is in choosing u(x) and v (x) correctly. The LIATE rule is a rule of thumb that tells you which function you should choose as u(x): LIATE The word itself tells you in which order of priority you should use u(x).
  • 14.
  • 15.
  • 16.
    Example 1 x sinxdx Here we have an algebraic and a trigonometric function:
  • 17.
    Example 1 x sinxdx Here we have an algebraic and a trigonometric function: x sin xdx
  • 18.
    Example 1 x sinxdx Here we have an algebraic and a trigonometric function:   A x sin xdx
  • 19.
    Example 1 x sinxdx Here we have an algebraic and a trigonometric function: x$$$XT sin xdx
  • 20.
    Example 1 x sinxdx Here we have an algebraic and a trigonometric function: x sin xdx
  • 21.
    Example 1 x sinxdx Here we have an algebraic and a trigonometric function: x sin xdx According to LIATE, the A comes before the T, so we choose u(x) = x.
  • 22.
  • 23.
  • 24.
    Example 2 ex sin xdx Inthis case we have E and T:
  • 25.
    Example 2 ex sin xdx Inthis case we have E and T: b E ex sin xdx
  • 26.
    Example 2 ex sin xdx Inthis case we have E and T: ex$$$XT sin xdx
  • 27.
    Example 2 ex sin xdx Inthis case we have E and T: ex sin xdx
  • 28.
    Example 2 ex sin xdx Inthis case we have E and T: ex sin xdx According to LIATE, the T comes before the E, so we choose u(x) = sin x.
  • 29.
    Example 2 ex sin xdx Inthis case we have E and T: ex sin xdx According to LIATE, the T comes before the E, so we choose u(x) = sin x. Let’s solve this integral to show that this works:
  • 30.
    Example 2 ex sin xdx Inthis case we have E and T: ex sin xdx According to LIATE, the T comes before the E, so we choose u(x) = sin x. Let’s solve this integral to show that this works: u(x) = sin x
  • 31.
    Example 2 ex sin xdx Inthis case we have E and T: ex sin xdx According to LIATE, the T comes before the E, so we choose u(x) = sin x. Let’s solve this integral to show that this works: u(x) = sin x v (x) = ex
  • 32.
    Example 2 ex sin xdx Inthis case we have E and T: ex sin xdx According to LIATE, the T comes before the E, so we choose u(x) = sin x. Let’s solve this integral to show that this works: u(x) = sin x v (x) = ex u (x) = cos x
  • 33.
    Example 2 ex sin xdx Inthis case we have E and T: ex sin xdx According to LIATE, the T comes before the E, so we choose u(x) = sin x. Let’s solve this integral to show that this works: u(x) = sin x v (x) = ex u (x) = cos x v(x) = ex
  • 34.
    Example 2 ex sin xdx Inthis case we have E and T: ex sin xdx According to LIATE, the T comes before the E, so we choose u(x) = sin x. Let’s solve this integral to show that this works: u(x) = sin x v (x) = ex u (x) = cos x v(x) = ex ex sin xdx = ex sin x − ex cos xdx
  • 35.
    Example 2 ex sin xdx Inthis case we have E and T: ex sin xdx According to LIATE, the T comes before the E, so we choose u(x) = sin x. Let’s solve this integral to show that this works: u(x) = sin x v (x) = ex u (x) = cos x v(x) = ex b v (x) ex sin xdx = ex sin x − ex cos xdx
  • 36.
    Example 2 ex sin xdx Inthis case we have E and T: ex sin xdx According to LIATE, the T comes before the E, so we choose u(x) = sin x. Let’s solve this integral to show that this works: u(x) = sin x v (x) = ex u (x) = cos x v(x) = ex ex$$$Xu(x) sin xdx = ex sin x − ex cos xdx
  • 37.
    Example 2 ex sin xdx Inthis case we have E and T: ex sin xdx According to LIATE, the T comes before the E, so we choose u(x) = sin x. Let’s solve this integral to show that this works: u(x) = sin x v (x) = ex u (x) = cos x v(x) = ex ex sin xdx = b v(x) ex sin x − ex cos xdx
  • 38.
    Example 2 ex sin xdx Inthis case we have E and T: ex sin xdx According to LIATE, the T comes before the E, so we choose u(x) = sin x. Let’s solve this integral to show that this works: u(x) = sin x v (x) = ex u (x) = cos x v(x) = ex ex sin xdx = ex$$$Xu(x) sin x − ex cos xdx
  • 39.
    Example 2 ex sin xdx Inthis case we have E and T: ex sin xdx According to LIATE, the T comes before the E, so we choose u(x) = sin x. Let’s solve this integral to show that this works: u(x) = sin x v (x) = ex u (x) = cos x v(x) = ex ex sin xdx = ex sin x − b v(x) ex cos xdx
  • 40.
    Example 2 ex sin xdx Inthis case we have E and T: ex sin xdx According to LIATE, the T comes before the E, so we choose u(x) = sin x. Let’s solve this integral to show that this works: u(x) = sin x v (x) = ex u (x) = cos x v(x) = ex ex sin xdx = ex sin x − ex $$$Xu (x) cos xdx
  • 41.
    Example 2 ex sin xdx Inthis case we have E and T: ex sin xdx According to LIATE, the T comes before the E, so we choose u(x) = sin x. Let’s solve this integral to show that this works: u(x) = sin x v (x) = ex u (x) = cos x v(x) = ex ex sin xdx = ex sin x − ex cos xdx
  • 42.
  • 43.
    Example 2 Now weneed to solve that integral:
  • 44.
    Example 2 Now weneed to solve that integral: ex sin xdx = ex sin x − ex cos xdx
  • 45.
    Example 2 Now weneed to solve that integral: ex sin xdx = ex sin x − ex cos xdx We need to use integration by parts again.
  • 46.
    Example 2 Now weneed to solve that integral: ex sin xdx = ex sin x − ex cos xdx We need to use integration by parts again. By using LIATE, we choose u(x) = cos x:
  • 47.
    Example 2 Now weneed to solve that integral: ex sin xdx = ex sin x − ex cos xdx We need to use integration by parts again. By using LIATE, we choose u(x) = cos x: u(x) = cos x
  • 48.
    Example 2 Now weneed to solve that integral: ex sin xdx = ex sin x − ex cos xdx We need to use integration by parts again. By using LIATE, we choose u(x) = cos x: u(x) = cos x v (x) = ex
  • 49.
    Example 2 Now weneed to solve that integral: ex sin xdx = ex sin x − ex cos xdx We need to use integration by parts again. By using LIATE, we choose u(x) = cos x: u(x) = cos x v (x) = ex u (x) = − sin x
  • 50.
    Example 2 Now weneed to solve that integral: ex sin xdx = ex sin x − ex cos xdx We need to use integration by parts again. By using LIATE, we choose u(x) = cos x: u(x) = cos x v (x) = ex u (x) = − sin x v(x) = ex
  • 51.
    Example 2 Now weneed to solve that integral: ex sin xdx = ex sin x − ex cos xdx We need to use integration by parts again. By using LIATE, we choose u(x) = cos x: u(x) = cos x v (x) = ex u (x) = − sin x v(x) = ex ex sin xdx = ex sin x − ex cos xdx
  • 52.
    Example 2 Now weneed to solve that integral: ex sin xdx = ex sin x − ex cos xdx We need to use integration by parts again. By using LIATE, we choose u(x) = cos x: u(x) = cos x v (x) = ex u (x) = − sin x v(x) = ex ex sin xdx = ex sin x − b v (x) ex cos xdx
  • 53.
    Example 2 Now weneed to solve that integral: ex sin xdx = ex sin x − ex cos xdx We need to use integration by parts again. By using LIATE, we choose u(x) = cos x: u(x) = cos x v (x) = ex u (x) = − sin x v(x) = ex ex sin xdx = ex sin x − ex $$$Xu(x) cos xdx
  • 54.
    Example 2 Now weneed to solve that integral: ex sin xdx = ex sin x − ex cos xdx We need to use integration by parts again. By using LIATE, we choose u(x) = cos x: u(x) = cos x v (x) = ex u (x) = − sin x v(x) = ex ex sin xdx = ex sin x − ex cos xdx
  • 55.
    Example 2 Now weneed to solve that integral: ex sin xdx = ex sin x − ex cos xdx We need to use integration by parts again. By using LIATE, we choose u(x) = cos x: u(x) = cos x v (x) = ex u (x) = − sin x v(x) = ex ex sin xdx = ex sin x − ex cos xdx = ex sin x − ex cos x − ex (− sin x) dx
  • 56.
    Example 2 Now weneed to solve that integral: ex sin xdx = ex sin x − ex cos xdx We need to use integration by parts again. By using LIATE, we choose u(x) = cos x: u(x) = cos x v (x) = ex u (x) = − sin x v(x) = ex ex sin xdx = ex sin x − ex cos xdx = ex sin x − b v(x) ex cos x − ex (− sin x) dx
  • 57.
    Example 2 Now weneed to solve that integral: ex sin xdx = ex sin x − ex cos xdx We need to use integration by parts again. By using LIATE, we choose u(x) = cos x: u(x) = cos x v (x) = ex u (x) = − sin x v(x) = ex ex sin xdx = ex sin x − ex cos xdx = ex sin x − ex $$$Xu(x) cos x − ex (− sin x) dx
  • 58.
    Example 2 Now weneed to solve that integral: ex sin xdx = ex sin x − ex cos xdx We need to use integration by parts again. By using LIATE, we choose u(x) = cos x: u(x) = cos x v (x) = ex u (x) = − sin x v(x) = ex ex sin xdx = ex sin x − ex cos xdx = ex sin x − ex cos x − b v(x) ex (− sin x) dx
  • 59.
    Example 2 Now weneed to solve that integral: ex sin xdx = ex sin x − ex cos xdx We need to use integration by parts again. By using LIATE, we choose u(x) = cos x: u(x) = cos x v (x) = ex u (x) = − sin x v(x) = ex ex sin xdx = ex sin x − ex cos xdx = ex sin x − ex cos x − ex $$$$$Xu (x) (− sin x)dx
  • 60.
    Example 2 Now weneed to solve that integral: ex sin xdx = ex sin x − ex cos xdx We need to use integration by parts again. By using LIATE, we choose u(x) = cos x: u(x) = cos x v (x) = ex u (x) = − sin x v(x) = ex ex sin xdx = ex sin x − ex cos xdx = ex sin x − ex cos x − ex (− sin x) dx
  • 61.
  • 62.
    Example 2 So, wehave that:
  • 63.
    Example 2 So, wehave that: ex sin xdx = ex sin x − ex cos x − ex sin xdx
  • 64.
    Example 2 So, wehave that: ex sin xdx !! = ex sin x − ex cos x − ex sin xdx !!
  • 65.
    Example 2 So, wehave that: ex sin xdx !! = ex sin x − ex cos x − ex sin xdx !! So, we add ex sin xdx to both sides of the equation:
  • 66.
    Example 2 So, wehave that: ex sin xdx !! = ex sin x − ex cos x − ex sin xdx !! So, we add ex sin xdx to both sides of the equation: 2 ex sin xdx = ex (sin x − cos x)
  • 67.
    Example 2 So, wehave that: ex sin xdx !! = ex sin x − ex cos x − ex sin xdx !! So, we add ex sin xdx to both sides of the equation: 2 ex sin xdx = ex (sin x − cos x) ex sin xdx = ex 2 (sin x − cos x) + C
  • 68.
    Example 2 So, wehave that: ex sin xdx !! = ex sin x − ex cos x − ex sin xdx !! So, we add ex sin xdx to both sides of the equation: 2 ex sin xdx = ex (sin x − cos x) ex sin xdx = ex 2 (sin x − cos x) + C This is in fact an example of Trick number 2.
  • 69.
  • 70.
    Example 3 Let’s doanother example:
  • 71.
    Example 3 Let’s doanother example: x2 ex dx
  • 72.
    Example 3 Let’s doanother example: x2 ex dx We have an A and an E:
  • 73.
    Example 3 Let’s doanother example: x2 ex dx We have an A and an E:    A x2 ex dx
  • 74.
    Example 3 Let’s doanother example: x2 ex dx We have an A and an E: x2 b E ex dx
  • 75.
    Example 3 Let’s doanother example: x2 ex dx We have an A and an E: x2 ex dx
  • 76.
    Example 3 Let’s doanother example: x2 ex dx We have an A and an E: x2 ex dx LIATE says that the A comes before the E. So, we choose u(x) = x2:
  • 77.
    Example 3 Let’s doanother example: x2 ex dx We have an A and an E: x2 ex dx LIATE says that the A comes before the E. So, we choose u(x) = x2: u(x) = x2
  • 78.
    Example 3 Let’s doanother example: x2 ex dx We have an A and an E: x2 ex dx LIATE says that the A comes before the E. So, we choose u(x) = x2: u(x) = x2 v (x) = ex
  • 79.
    Example 3 Let’s doanother example: x2 ex dx We have an A and an E: x2 ex dx LIATE says that the A comes before the E. So, we choose u(x) = x2: u(x) = x2 v (x) = ex u (x) = 2x
  • 80.
    Example 3 Let’s doanother example: x2 ex dx We have an A and an E: x2 ex dx LIATE says that the A comes before the E. So, we choose u(x) = x2: u(x) = x2 v (x) = ex u (x) = 2x v(x) = ex
  • 81.
    Example 3 Let’s doanother example: x2 ex dx We have an A and an E: x2 ex dx LIATE says that the A comes before the E. So, we choose u(x) = x2: u(x) = x2 v (x) = ex u (x) = 2x v(x) = ex x2 ex dx =
  • 82.
    Example 3 Let’s doanother example: x2 ex dx We have an A and an E: x2 ex dx LIATE says that the A comes before the E. So, we choose u(x) = x2: u(x) = x2 v (x) = ex u (x) = 2x v(x) = ex    u(x) x2 ex dx =
  • 83.
    Example 3 Let’s doanother example: x2 ex dx We have an A and an E: x2 ex dx LIATE says that the A comes before the E. So, we choose u(x) = x2: u(x) = x2 v (x) = ex u (x) = 2x v(x) = ex x2 b v (x) ex dx =
  • 84.
    Example 3 Let’s doanother example: x2 ex dx We have an A and an E: x2 ex dx LIATE says that the A comes before the E. So, we choose u(x) = x2: u(x) = x2 v (x) = ex u (x) = 2x v(x) = ex x2 ex dx = x2 ex − 2xex dx
  • 85.
    Example 3 Let’s doanother example: x2 ex dx We have an A and an E: x2 ex dx LIATE says that the A comes before the E. So, we choose u(x) = x2: u(x) = x2 v (x) = ex u (x) = 2x v(x) = ex x2 ex dx =    u(x) x2 ex − 2xex dx
  • 86.
    Example 3 Let’s doanother example: x2 ex dx We have an A and an E: x2 ex dx LIATE says that the A comes before the E. So, we choose u(x) = x2: u(x) = x2 v (x) = ex u (x) = 2x v(x) = ex x2 ex dx = x2 b v(x) ex − 2xex dx
  • 87.
    Example 3 Let’s doanother example: x2 ex dx We have an A and an E: x2 ex dx LIATE says that the A comes before the E. So, we choose u(x) = x2: u(x) = x2 v (x) = ex u (x) = 2x v(x) = ex x2 ex dx = x2 ex − b u (x) 2xex dx
  • 88.
    Example 3 Let’s doanother example: x2 ex dx We have an A and an E: x2 ex dx LIATE says that the A comes before the E. So, we choose u(x) = x2: u(x) = x2 v (x) = ex u (x) = 2x v(x) = ex x2 ex dx = x2 ex − 2xb v(x) ex dx
  • 89.
    Example 3 Let’s doanother example: x2 ex dx We have an A and an E: x2 ex dx LIATE says that the A comes before the E. So, we choose u(x) = x2: u(x) = x2 v (x) = ex u (x) = 2x v(x) = ex x2 ex dx = x2 ex − 2xex dx
  • 90.
  • 91.
    Example 3 x2 ex dx =x2 ex − 2 xex dx
  • 92.
    Example 3 x2 ex dx =x2 ex − 2 xex dx Now we need to solve that integral by integration by parts again:
  • 93.
    Example 3 x2 ex dx =x2 ex − 2 xex dx Now we need to solve that integral by integration by parts again: u(x) = x
  • 94.
    Example 3 x2 ex dx =x2 ex − 2 xex dx Now we need to solve that integral by integration by parts again: u(x) = x v (x) = ex
  • 95.
    Example 3 x2 ex dx =x2 ex − 2 xex dx Now we need to solve that integral by integration by parts again: u(x) = x v (x) = ex u (x) = 1
  • 96.
    Example 3 x2 ex dx =x2 ex − 2 xex dx Now we need to solve that integral by integration by parts again: u(x) = x v (x) = ex u (x) = 1 v(x) = ex
  • 97.
    Example 3 x2 ex dx =x2 ex − 2 xex dx Now we need to solve that integral by integration by parts again: u(x) = x v (x) = ex u (x) = 1 v(x) = ex x2 ex dx = x2 ex − 2 xex − 1.ex dx
  • 98.
    Example 3 x2 ex dx =x2 ex − 2 xex dx Now we need to solve that integral by integration by parts again: u(x) = x v (x) = ex u (x) = 1 v(x) = ex x2 ex dx = x2 ex − 2 xex − 1.ex dx = x2 ex − 2xex + 2ex = ex x2 − 2x + 2
  • 99.
    Example 3 x2 ex dx =x2 ex − 2 xex dx Now we need to solve that integral by integration by parts again: u(x) = x v (x) = ex u (x) = 1 v(x) = ex x2 ex dx = x2 ex − 2 xex − 1.ex dx = x2 ex − 2xex + 2ex = ex x2 − 2x + 2