Hi
I need help on the following questions:
1)
This is what i have done.
u = x du = 1 dx
dv = sin(2x) dx
final answer.
2)
This is what i have done, not sure if it is correct.
P.S
There should not be an integral sign between "5" and "sin(2x)/2".
Yes, this is correct. You can simplify it slightly by writing cos(-3x)= cos(3x) since cosine is an even function. Of course, in your final solution, -sin(-3x)= -(- sin(3x))= sin(3x) since sine is an odd function.u = x du = 1 dx
dv = sin(2x) dx
final answer.
2)
This is what i have done, not sure if it is correct.
P.S
It would be an interesting exercise in trigonometric identities to show that your solution is the same as Prove It's solution (I did not check his work). Although your solution is considerably easier than his!