Page 81 - DMTH202_BASIC_MATHEMATICS_II

Basic Mathematics-II

Notes The limits are a little strange in this case, but that will occur sometimes so don’t get too
energized regarding it. Here is the substitution.
x

u  1 e 2 du   e dx
0

x  0  u  1 e  1 1 0




x  In 1     u  1 e In 1   1 1   
The integral is then,
In 1  x 2 
0  e cos 1 e  dx   0  cos u du

  sin   u
0
  sin   sin 0   0
6  4
e In t
(b) 2  dt
e t
Here is the substitution and converted limits for this problem.
1
u  In t du  dt
t
2
6
t  e 2  u  In e  2 t  e 6  u  In e  6
The integral is,
e 6 In t  4 6 4
 dt  2  u du
t
6
1
 u 5
5 2
7744

5
 sec 3P tan 3P 
9 dP
(c)   3

12 2 sec 3P 
Here is the substitution and converted limits and don’t get too thrilled about the
substitution. It’s a little chaotic in the case, but that can take place on occasion.
1


P
P
u  2  sec(3 ) du  3 sec (3 ) tan (3 )dP  sec 3P tan 3P dP du
P
3
   

P   u  2 sec    2  2
12  4 
   
P   u  2  sec    4
9  3 
Here is the integral,
 sec 3P tan 3P  1 4  1
9 3
  dP  2  u du

3 2 sec 3P  3 2
12
4
2
1
 u 3
2
2 2
1  2 2 
  4   2   2 3 
3
2  
So, not only was the substitution chaotic, but we also a untidy answer, but again that’s life
on occasion.
76 LOVELY PROFESSIONAL UNIVERSITY