Page 19 - DMTH202_BASIC_MATHEMATICS_II

Basic Mathematics-II

Notes Thus,
2
 sec x dx
Now we locate the definite integral

2
4
x
4
 xe dx  1 e x 2 2 0  1 (e  e 0 )  1 (e  1) .
2 2 2
There is another method observe at the same problem. After we recognized that
1 w
2
x
 xe dx  e  C ,
2
2
our next two steps were to substitute w by x , and then x by 2 and 0. We could have directly
substituted the original limits of integration x = 0, and x = 2, by the analogous w limits. As
2
2
2
w = x , the w limits are w = 0 = 0 (when x = 2 ) and w = 2 = 4 (when x = 2), so we get
x 2 x 2 1 w 4 w 1 w 4 1 4 0 1 4
0  xe dx  w  e dx  e  (e  e ) (e  1).
x 2 0 2 0 2 2
As we would guess, both methods provide the similar answer.

3
Task Evaluate x 2 1 x dx by using substitution.


1.2.5 More Complex Substitutions

In the examples of substitution illustrated until now, we guessed an expression for w and
expected to find (or some constant multiple of it) in the integrand. What if we are not so
fortunate? It turns out that it frequently works to let w be some muddled expression contained
inside, say, a cosine or under a root, even if we cannot see instantly how such a substitution
assists.

Example: Find  1  xdx .
Solution:
Here, the derivative of the inside function is nowhere to be observed. Yet, we try w   x .
1
Then
2
w  1  x , so  – 1w   , x so.
Thus 2(w – 1 dw = dx).
We have

 1  x dx   w 2(w  2)dw  2 w 1/2 (w  1)

 2 (w 3/2  w 1/2 )dx  2   2 w 5/2  2 w 3/2    C

 5 3 
 2 5/2 2 3/2 
 2   1  x    1  x    C
 5 3 

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