Page 17 - DMTH202_BASIC_MATHEMATICS_II

Basic Mathematics-II

Notes w
2  1) (t 2  1) w 1 1 w 1 w 1 (t 2  1)
(t
 t e dt – e tdt – e dw –  e dw – e  C – e  C
 

1 2 2 2 2
dw
2
This provides the similar answer as we found by means of guess-and-check.
1 w 1 w 1
Why didn’t we put x  e dw – e  C in the preceding example? As the constant C is arbitrary,
2 2 2
1
it doesn’t actually matter whether we add C or C . The convention is always to add C to
2
whatever antiderivative we have computed.
Now let’s do again the third example that we solved formerly by guess-and-check.

Example: Find x 3 x 4  5 dx .

Solution:
The inside function is x +5, with derivative 4x . The integrand has a factor of x , and as the only
3
3
4
thing missing is a constant factor, we attempt
4
w = x + 5
Then
3
dw  w '( )dx  4x dx
x
Giving
1 3
aw  x dx
4
So,

1 1 1/2 1 w 3/2 1 4 3/2
3
4
 x x  5 dx –  w dw –  w dw – .  C – (x  5) –C
4 4 4 3/2 6
Again, we get the similar result as with guess-and-check.
!
Caution We can apply the substitution method when a constant factor is absent from the
derivative of the inside function. Though, we may not be able to utilize substitution if
anything other than a constant factor is missing. For example, setting w = x – 5 to find
4
 x 2 x 4  5 dx
2
3
does us no good since x dx is not a constant multiple of dw=4x dx. Substitution functions
if the integrand encloses the derivative of the inside function, to within a constant factor.
Some people favor the substitution method over guess-and-check as it is more systematic, but
both methods attain the same result. For uncomplicated problems, guess-and-check can be quicker.

d

Example: Find e cos  sin  .
Solution:
Consider w = cos  as its derivative is –sin and there is a factor of sin in the integrand. This
provides

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