Integration by Parts: “Undoing” the Product Rule for Derivatives  Look at the derivative of a product of functions: d dv du uv = u +v ( ) dx dx dx  Let’s use the differential form: d (uv ) =udv+ vdu  And solve for udv udv =d (uv) −vdu  Integrating both sides, we get: ∫udv = ∫d (uv) −∫vdu
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Integration by Parts: “Undoing” the Product Rule for Derivatives  Integrating both sides, we get: ∫udv = ∫d (uv) −∫vdu    Or ∫udv =uv−∫vdu ∫vdu should be simpler that the original ∫udv The integral If two functions are not related by derivatives (substitution does not apply), choose one function to be the u (to differentiate) and the other function to be the dv (to integrate)
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Integration by Parts ∫udv =uv−∫vdu Integrate: ∫cos (ln(x))dx ∫cos (ln(x))dx =x cos ( ln(x)) + ∫ sin ( ln(x)) dx u =sin (ln(x)) 1 du = cos(ln(x))dx x dv =dx v =x ∫cos (ln(x))dx =x cos ( ln(x)) +x sin ( ln(x)) −∫ cos ( ln(x)) dx + ∫ cos ( ln(x)) dx + ∫ cos ( ln(x)) dx 2 ∫cos (ln(x))dx =x cos ( ln(x)) + x sin ( ln(x)) 1 cos ln(x) dx = ( ) ( x cos ( ln(x)) + x sin ( ln(x))) ∫ 2
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8.1 Integration by parts d dv du  uv  u  v Product Rule: dx dx dx d dv du dx  uv dx u dxdx  v dx dx uv udv  vdu udv  uv  vdu  
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Integration by Parts d(uv) = udv + vdu udv = uv - vdu Show that xnexdx = xnex - nxn-1exdx + C let u = xn; dv = exdx then du = nxn-1dx; v = ex + C Thus, xnexdx = xnex - nxn-1ex dx + C 03/22/2019 rd 12
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Θ(θ) Part 2  Let l=0: d  d d  d d   P0 (cos )   (1)    sin  (0)  0  sin   sin  d  d d  d  d      0(0  1) P0 (cos  ) sin( ) 0 0 0  Let l=3: d  d  d P3 (cos  )    sin  d  d  d ↘  d d  ↘  3     d 2 3   sin  d  1   (cos   1 )    d  2 3 3!  d cos( )       d 5 3 d  3   15  3  sin   sin    cos 2  sin   sin     cos   cos    d  2 2 d  2 2       ↘ 3  15   3(3  1) P3 (cos  )  12  cos 2  sin   sin   2  2  □ d  15 2 3 2  2 3 3   cos  sin   sin    15(cos  sin   cos sin  )  3 sin  cos d  2 2  ↓   18  5 cos 2  sin   sin   15(cos 3  sin   cos  sin 3  )  3 sin  cos  □
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Integration by parts 2 u = x2 dv = sin x dx du = 2x dx v = -cos x x sin xdx  x 2 udv uv  vdu 2 sin xdx  x cos  2 x cos xdx u = 2x dv = cos x dx du = 2dx v = sin x 2 2 x sin xdx  x cos x  2 x sin x  2sin xdx  x 2 2 sin xdx  x cos x  2 x sin x  2cos x  C
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Evaluating F(s)=L{f(t)}- Hard Way udv uv  vdu remember let u e  st , du  se  st dt dv sin( t )dt , v  cos( t )     st  st  e sin( t )dt [ e cos(t ) ]  s e  st cos(t )dt  0 0 0   se  e (1)  s e  st cos(t )dt  1  st e sin( t ) dt  2  1  s 0  e  st cos(t )dt  0 [ e  sin(t ) ]  s e 0 0 0 0    st  st e  sin( t )dt  (1  s 2 ) e  st sin( t )dt 1 u e  st , du  se  st dt dv cos(t )dt, v sin( t )  st sin( t )dt 1  s 2  0    st 0  st let Substituting, we get:  st sin(t )dt  e (0)  s e 0  st sin(t )dt It only gets worse…
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Navigation Mathematics : Kinematics – Rotation Matrices • In general, Rotations do NOT commute  Consider R( yˆ , ) R( zˆ , ) R( zˆ , ) R( yˆ , ) Tuesday 22 Jan 2013  Cos( ) 0 Sin( )   Cos( )  Sin( ) 0   0 1 0   Sin( ) Cos( ) 0    Sin( ) 0 Cos( )   0 0 1   Cos( )Cos( )  Cos( )Sin( ) Sin( )   Sin( ) Cos( ) 0    Sin( )Cos( ) Sin( )Sin( ) Cos( )   Cos( )Cos( )  Sin( ) Cos( )Sin( )   Sin( )Cos( ) Cos( ) Sin( )Sin( )  R( yˆ , ) R( zˆ , )   Sin( ) 0 Cos( )  NMT EE 570: Location and Navigation: Theory & Practice Slide 6 of 20
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Integration by Parts Integrate: ∫cos (ln(x))dx ∫udv =uv−∫vdu
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Integration by Parts Back to:  ∫udv =uv−∫vdu ∫x ln(x)dx Choose u (to differentiate (“du”)) dv (to integrate (“v”)) u =ln(x) 1 du = dx x dv = xdx x2 v= 2 x2 ⎛ x 2⎞⎞ ⎛ 1 ⎞ ∫x ln(x)dx = 2 ln(x) −∫ ⎜⎝ 22⎟⎠⎟⎠dx⎜⎝ x ⎟⎠ dx x2 x2 = ln(x) − + C 2 4 This second integral is simpler than the first
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Integration by Parts Evaluate:  y 3 5 ∫y e ∫udv =uv−∫vdu dy Choose u (to differentiate (“du”)) dv (to integrate (“v”)) y 5 u =y 3 y 3 5 y 2 5 dv =ye dy 3y 2 5e 5 6y 25e 6 125e 0 625e y 5 y 5 y 5 y 5 y 5 y 5 5y e −75 y e + 750 ye −3750 e =e ( 5y 3 − 75y 2 + 750y − 3750 )
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Integration by Parts Evaluate:  y 3 5 ∫y e ∫udv =uv−∫vdu dy Choose u (to differentiate (“du”)) dv (to integrate (“v”)) Chapter 7.1 March 22, 2007
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Integration by Parts ∫udv =uv−∫vdu 0 Evaluate:  ∫(2x + 6)e −x dx −1 Choose u (to differentiate (“du”)) dv (to integrate (“v”))
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Powers of sin and cos 3 2 2 sin (2  ) d   sin 2  sin 2  d   (1  cos 2 )sin 2 d    1 1 3 (sin 2   cos 2  sin 2  ) d    cos 2   cos 2  C  2 6 2 2 3 2 2 2 2 sin  cos  d   sin  cos  cos  d   sin  (1  sin  )cos  d    2 3 2 3 (sin   sin  ) cos  d   (sin  cos   sin  cos  ) d   1 3 1 4 sin   sin   C 3 4
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Integration by Parts ∫udv =uv−∫vdu Examples: 3 x2 ∫x e dx ∫tan x ln(cos x)dx x x e cos(e )dx ∫ x e ∫ cos(4 x)dx
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Or you can use Integration by Parts ⎛t ⎞ ∫cos ⎜⎝ 2 ⎟⎠ dx 2 ∫udv =uv−∫vdu
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DCAF From Executor From Scheduler M&C CMCS Correlation Results CBE Archive Pointers to Correlation Results DCAF Time Metadata Metadata Metadata Integration 1 Integration 2 ... Integration m Integration 1 Integration 2 ... Integration m Integration 1 Integration 2 ... SubScan 1 SubScan 2 Scan 1 SubScan 3 Exec Block 1 Integration m ... ... ... ... Integration 1 Integration 2 ... Metadata SubScan M Scan N Integration m 14 June, 2004 EVLA Overall Design Subsystems II Tom Morgan 21
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