Page 15 - DMTH202_BASIC_MATHEMATICS_II

Basic Mathematics-II

Notes Solution:
2
Since the radical has the form a  x 2
let x = a sin = 2 sin
dx = 2 cosd

and 4 x 2  2cos

dx 2cos d


hence, 2  2  2
x 4 x  4sin  2cos 

1 d
 
2
4 sin 
1
2
  sec  d
4
1
 cot  C
4
1 4 x 2

  C
4 x
4 x 2

  C
4x
dx
Example: Evaluate 

25 x 2
Solution:
2
Since the radical has the form a  x 2
let x = a tan = 5 sin
dx = 5 sec d
2
and 25 x 2  5cos

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