Page 65 - DMTH202_BASIC_MATHEMATICS_II

Basic Mathematics-II

Notes
1 A B
 
(x  1)(x  2) x  1 x  2

 1  A (x  2) B (x  1)
Putting x = – 1, we get A = 1

Putting x = – 2, we get B = – 1
2 dx 2  1 1 

I  1  (x  1)(x  2)  1    x  1 x    dx
2
 log|x  1| log|x  2|  2

1
3 2 9


 (log 3 log 4) (log 2 log 3) log  log  log


4 3 8
3 dx
(ii) I   x 2 (x  1)
0
1 A B C
  
x 2 (x  1) x x 2 x  1

 1  Ax (a  1) B (x  1) Cx 2

Putting x = – 1, we get C = 1
Putting x = 1, we get A = – 1
3 dx 3  1 1 1 
I  1  x 2 (x  1)  1      x 2  x  1  dx
x
3
1 
x
  log| |  log|x  1|
x  0 
 1   1 
  log 3   log 4   log|1|  log|2| 

 
 3   1 
1 2 2
  log 3   log 4 1 log 2   log


3 3 3
Example: Evaluate the following integrals:
 /4
x
(i) 0  sin2 cos3xdx
 /4
3
(ii) 0  cos xdx
 /4
(iii) 0  1 sin 2x dx

 /2
(iv) 0  1 cosxdx

 sin x
(v) 0  sin x  cosx dx
Solution:
/4
x
(i) I  sin2 cos3xdx
 0

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