Page 18 - DMTH202_BASIC_MATHEMATICS_II

Unit 1: Integration

dw = w’ (d= –sind Notes
So
–dw = sind
Therefore

w

d
 e cos  sin  – e w (–dw )–(–1) e dw – e w  C   e cos  C


e t
Example: Find  dt .
1  e t
Solution:
t
t
t
Observing that the derivative of 1 + e is e , we notice w = 1 + e is a good quality choice. Then
t
dw = e dt, so that
e t 1 t 1
 t dt   t e dt   dw  1n w  C
1  e 1  e w
n
 1 1  e t C
As the numerator is e dt, we might also have attempted w = e . This substitution leads to the
t
t

integral 1/(1 w dw  ) , which is better than the original integral but needs another substitution,
u = 1 + w, to terminate. There are frequently several different methods of doing an integral by
substitution.

Notes Observe the pattern in the preceding example: having a function in the denominator
and its derivative in the numerator shows a natural logarithm. The next example follows
the similar pattern.


Example: Find tan d .
Solution:
Remember that tan =(sin   (cos  . If w = cos, then dw = –sin d, so

sin dw
n
 tan d =  d =    1 cos  C

cos w
2 2
x
Example: Compute 0  xe dx
Solution:
To assess this definite integral by means of the Fundamental Theorem of Calculus, we first want
2
2
to find an antiderivative of ( )f x  xe x 2 . The inside function is x , so we consider w = x . Then
dx = 2x dx, so
1
dx  xdx .
2

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