Page 27 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION
P. 27
Differential and Integral Equation
Notes Multiplying both sides of (i) by cos 2m and then integrating between the limits 0 to , we get
cos( sin )cos 2m d
x
0
2
= J 0 cos2m d 2J 2 cos2 cos2m d ... 2J 2m cos 2m d ...
0 0 0
= 0 + 0 + ... + J 2m (1 cos4m )d ...
0
= J .
2m
Similarly, we can prove that
cos( sin )cos(2m 1) d 0
x
0
Again multiplying both sides of (ii) by sin (2m + 1) and then integrating between the limits 0 to
, we get
sin( sin )sin(2m 1) d
x
0
= 2J 1 sin .sin(2m 1) d 2J 2 sin3 sin(2m 1) d
0 0
2
... 2J sin (2m 1) d ...
2m 1
0
= 0 + 0 +...+ 2J 2m 1 1 cos2(2m 1) d ...
0
= J [ ] J
2m+1 0 2m 1
Similarly,
x
sin( sin )sin2m d 0
0
Therefore
m
x
cos(2m x sin )d cos2 .cos( sin )d
0 0
x
+ sin2m .sin( sin )d
0
= J
2m
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