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Integration by parts udv uv vdu d dv du uv u v dx dx dx Product Rule: d dv du uv dx u dx v dx dx dx dx uv udv vdu

Integration by parts Product Rule:. Integration by parts Let dv be the most complicated part of the original integrand that fits a basic integration Rule

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Page 1: Integration by parts Product Rule:. Integration by parts Let dv be the most complicated part of the original integrand that fits a basic integration Rule

Integration by parts

udv uv vdu

d dv duuv u v

dx dx dx Product Rule:

d dv duuv dx u dx v dx

dx dx dx

uv udv vdu

Page 2: Integration by parts Product Rule:. Integration by parts Let dv be the most complicated part of the original integrand that fits a basic integration Rule

Integration by parts

udv uv vdu Let dv be the most complicated part of theoriginal integrand that fits a basic integrationRule (including dx). Then u will be the remaining factors.

Let u be a portion of the integrand whose derivative is a function simpler than u. Thendv will be the remaining factors (including dx).

OR

Page 3: Integration by parts Product Rule:. Integration by parts Let dv be the most complicated part of the original integrand that fits a basic integration Rule

Integration by parts

udv uv vdu xxe dx u = x dv= exdx

du = dx v = ex

x x xxe dx xe e dx x x xxe dx xe e C

Page 4: Integration by parts Product Rule:. Integration by parts Let dv be the most complicated part of the original integrand that fits a basic integration Rule

Integration by parts

udv uv vdu 2 lnx xdx u = lnx dv= x2dx

du = 1/x dx v = x3 /3

3 3 3 22 1

ln ln ln3 3 3 3

x x x xx xdx x dx x dx

x

3 32 ln ln

3 9

x xx xdx x C

Page 5: Integration by parts Product Rule:. Integration by parts Let dv be the most complicated part of the original integrand that fits a basic integration Rule

Integration by parts

udv uv vdu arcsin xdx u = arcsin x dv= dx

2arcsin arcsin

1

xxdx x x dx

x

v = x2

1

1du dx

x

2arcsin arcsin 1xdx x x x C

Page 6: Integration by parts Product Rule:. Integration by parts Let dv be the most complicated part of the original integrand that fits a basic integration Rule

Integration by partsudv uv vdu

2 sinx xdx u = x2 dv = sin x dx

2 2sin cos 2 cosx xdx x x xdx du = 2x dx v = -cos x

u = 2x dv = cos x dx

du = 2dx v = sin x2 2sin cos 2 sin 2sinx xdx x x x x xdx

2 2sin cos 2 sin 2cosx xdx x x x x x C

Page 7: Integration by parts Product Rule:. Integration by parts Let dv be the most complicated part of the original integrand that fits a basic integration Rule

8.2 Trigonometric Integrals Powers of Sine and Cosine

sin cosn mu udusin cos cos sinn nu udu u udu

2 2sin 1 cosu u 1. If n is odd, leave one sin u factor and use

for all other factors of sin.

2 2cos 1 sinu u 2. If m is odd, leave one cos u factor and use

for all other factors of cos.

2 21sin (1 cos

1cos (1 cos2 ) 2 )

22oru u uu

3. If neither power is odd, use power reducing formulas:

Page 8: Integration by parts Product Rule:. Integration by parts Let dv be the most complicated part of the original integrand that fits a basic integration Rule

Powers of sin and cos

2 2sin ( )cos ( )d

3sin (2 )d

2 3sin ( )cos ( )d

Page 9: Integration by parts Product Rule:. Integration by parts Let dv be the most complicated part of the original integrand that fits a basic integration Rule

Powers of sin and cos3 2 2sin (2 ) sin 2 sin 2 (1 cos 2 )sin 2d d d

2 3 2 2 2 2sin cos sin cos cos sin (1 sin )cosd d d

2 3(sin 2 cos 2 sin 21 1

cos 2 cos2 6

) 2 Cd

2 3 2 3(sin sin )cos (sin cos sin cos )d d 3 41 1

sin sin3 4

C

Page 10: Integration by parts Product Rule:. Integration by parts Let dv be the most complicated part of the original integrand that fits a basic integration Rule

Powers of sin and cos

2 2 1 1sin ( )cos ( ) (1 cos 2 ) (1 cos 2 )

2 2d d

21 1 1(1 cos 2 ) (1 (1 cos 4 )

4 4 21 1 1 1

(1 cos 4 ) ( 4 )4 2 4 2

d d

d cos d

1 1sin 4

8 16C