Page 79 - DMTH202_BASIC_MATHEMATICS_II

Basic Mathematics-II

Notes 1 y 1 y 1
0  2 e  2cos   y dy  0  2 e dy  0  2 2 cos   y dy
Here is the substitution and converted limits for the second term.
1
u   y du   dy  dy  du

1 
y  0  u  0 y   u 
2 2
Here is the integral.
1 1 2 
y
y
0  2 e  2cos   y dy  0  2 e dy   0  2 cos   u du

1
2 2 2
y
 e  sin u

0 0
1 2  2
0
2
 e  e  sin  sin 0
 2 
1 2
 e  1 
2

0  z 
(d)   3sin    5cos    z dz
3  2 
This integral will need two substitutions. So first divide the integral so we can accomplish
a substitution on each term.
0  z  0  z  0
  3sin    5cos    z dz    3sin   dz    5cos    z dz
3  2  3  2  3
There are the two substitutions for these integrals.
z 1
u  du  dz  dz  2du
2 2
 
z   u  z  0  u 0
3 6
v    z dv   dz  dz   dv
 2
z   v  z  0  v 
3 3
Here is the integral for this problem.
0  z  0 
 
 
  3sin    5cos    z dz  6 sin   u du  5 2 cos   v dv
3  2  6 3
0 x
  6cos   5sinu    2x
v
6 3
 5 3 
 3 3 6   

 2 
3
  6
2
Example: Evaluate each of the following.
5 4t
(a) 5  2 dt
 2 8t


74 LOVELY PROFESSIONAL UNIVERSITY