3.2 Integration by Substitution

3.2 Integration by Substitution
1. It is given that ( ax+b ) n dx,n1.  

(A) Using the Substitution method,
Let u=ax+b Thus,  du dx =a        dx= du a

Example 1:
( 3x+5 ) 3 dx. Let u=3x+5     du dx =3     dx= du 3 ( 3x+5 ) 3 dx = u 3 du 3    substitute 3x+5=u and dx= du 3 = 1 3 u 3 du = 1 3 ( u 4 4 )+c = 1 3 ( ( 3x+5 ) 4 4 )+c    substitute back  u=3x+5   = ( 3x+5 ) 4 12 +c

(B) Using Formula method
( ax+b ) n = ( ax+b ) n+1 ( n+1 )a +c Hence,  ( 3x+5 ) 3 dx = ( 3x+5 ) 4 4( 3 ) +c                    = ( 3x+5 ) 4 12 +c


Example 2 (Formula method):

(a)  2 7 ( 5x ) 4 dx (b)  2 3 ( 9x2 ) 5 dx 

Solution:
(a)  2 7 ( 5x ) 4 dx= 2 ( 5x ) 3 7( 3 )( 1 ) +C                                      = 2 21 ( 5x ) 3 +C (b)  2 3 ( 9x2 ) 5 dx = 2 ( 9x2 ) 5 3 dx = 2 ( 9x2 ) 4 3( 4 )( 9 ) +C = 2 108 ( 9x2 ) 4 +C = 1 54 ( 9x2 ) 4 +C