Page 10 - DMTH202_BASIC_MATHEMATICS_II

Unit 1: Integration

d 3 3 2 Notes
(sin(x ))  cos(x ).(3x )
dx
As this is what we started with, we know that
3
2
3

 3x cos(x )dx–sin(x ) C.
The fundamental thought of this method is to attempt to locate an inside function whose derivative
occurs as a factor.
This functions even when the derivative is missing a constant factor, as in the subsequent
example.

2  1)
(t
Example: Find t e dt .

Solution:
2
It appears like t +1 is an inside function. So we guess e (t 2  1) for the anti derivative, as taking the
derivative of exponential outcomes in the recurrence of the exponential jointly with other terms
from the chain rule.
Now we check:
d (t 2  1) (t 2  1)
(e ) – (e ) 2t .
dt
1 (t 2  1)
The original guess was excessively large by a factor of 2. We modify the guess to e and
2
check yet again:

1
d   e (t 2  1) – 1 e (t 2  1) 2 – e (t 2  1) t
t
dt  2   2
Therefore, we know that

 t e (t 2  1) dt  1 e (t 2  1) –C
2

Example: Find x 3 x 4  5 dx

Solution:
4
At this point the inside function is x + 5, and its derivative occurs as a factor, with the exemption
of a missing 4. Therefore, the integrand we have is more or less of the form

x
g '( ) g ( )
x
with g(x) = x + 5. As x 3/2 /(3/2) is an anti derivative of the outside function x , we might guess
4
that an anti derivative is
g
( ( )) 3/2 (x 4  5) 3/2
x

3/2 3/2
Let’s check:
d (  x 4  5) 3/2  3 (x 4  5) 1/2 3 3 4 1/2
x
   .4x  4 (x  5)
dx  3/2  2 3/2

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