The LIATE Rule The LIATE Rule The difficulty of integration by parts - PowerPoint PPT Presentation

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 IATE ���� Log

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 ATE ���� Inv. Trig

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 TE ���� Algebraic

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 E ���� Trig

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: � ✘ ✿ T x ✘✘ 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 � e x sin xdx

Example 2 � e x sin xdx In this case we have E and T:

Example 2 � e x sin xdx In this case we have E and T: � ❃ E ✚ e x sin xdx ✚

Example 2 � e x sin xdx In this case we have E and T: � ✘ ✿ T e x ✘✘ sin xdx

Example 2 � e x sin xdx In this case we have E and T: � e x sin xdx

Example 2 � e x sin xdx In this case we have E and T: � e x sin xdx According to LIATE, the T comes before the E, so we choose u ( x ) = sin x .

Example 2 � e x sin xdx In this case we have E and T: � e x 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 � e x sin xdx In this case we have E and T: � e x 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 � e x sin xdx In this case we have E and T: � e x 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 ) = e x

Example 2 � e x sin xdx In this case we have E and T: � e x 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 ) = e x u ′ ( x ) = cos x

Example 2 � e x sin xdx In this case we have E and T: � e x 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 ) = e x u ′ ( x ) = cos x v ( x ) = e x

Example 2 � e x sin xdx In this case we have E and T: � e x 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 ) = e x u ′ ( x ) = cos x v ( x ) = e x � � e x sin xdx = e x sin x − e x cos xdx

Example 2 � e x sin xdx In this case we have E and T: � e x 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 ) = e x u ′ ( x ) = cos x v ( x ) = e x ❃ v ′ ( x ) � � ✚ e x sin xdx = e x sin x − e x cos xdx ✚

Example 2 � e x sin xdx In this case we have E and T: � e x 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 ) = e x u ′ ( x ) = cos x v ( x ) = e x � � ✿ u ( x ) ✘ e x ✘✘ sin xdx = e x sin x − e x cos xdx

Example 2 � e x sin xdx In this case we have E and T: � e x 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 ) = e x u ′ ( x ) = cos x v ( x ) = e x ❃ v ( x ) � � ✚ e x sin xdx = ✚ e x sin x − e x cos xdx

Example 2 � e x sin xdx In this case we have E and T: � e x 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 ) = e x u ′ ( x ) = cos x v ( x ) = e x � � ✿ u ( x ) ✘ e x sin xdx = e x ✘✘ e x cos xdx sin x −

Example 2 � e x sin xdx In this case we have E and T: � e x 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 ) = e x u ′ ( x ) = cos x v ( x ) = e x ❃ v ( x ) � � ✚ e x sin xdx = e x sin x − e x cos xdx ✚

Example 2 � e x sin xdx In this case we have E and T: � e x 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 ) = e x u ′ ( x ) = cos x v ( x ) = e x � � ✿ u ′ ( x ) ✘ e x sin xdx = e x sin x − e x ✘✘ cos xdx

Example 2 � e x sin xdx In this case we have E and T: � e x 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 ) = e x u ′ ( x ) = cos x v ( x ) = e x � � e x sin xdx = e x sin x − e x cos xdx

Example 2

Example 2 Now we need to solve that integral:

Example 2 Now we need to solve that integral: � � e x sin xdx = e x sin x − e x cos xdx

Example 2 Now we need to solve that integral: � � e x sin xdx = e x sin x − e x cos xdx We need to use integration by parts again.

Example 2 Now we need to solve that integral: � � e x sin xdx = e x sin x − e x 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: � � e x sin xdx = e x sin x − e x 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: � � e x sin xdx = e x sin x − e x 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 ) = e x

Example 2 Now we need to solve that integral: � � e x sin xdx = e x sin x − e x 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 ) = e x u ′ ( x ) = − sin x

Example 2 Now we need to solve that integral: � � e x sin xdx = e x sin x − e x 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 ) = e x u ′ ( x ) = − sin x v ( x ) = e x

Example 2 Now we need to solve that integral: � � e x sin xdx = e x sin x − e x 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 ) = e x u ′ ( x ) = − sin x v ( x ) = e x � � e x sin xdx = e x sin x − e x cos xdx