R Ryan Rigdon Junior Member Joined Jun 10, 2010 Messages 246 Sep 20, 2010 #1 I am stuck here. been on the problem for over an hour cant see what to do next. here is my work thus far.
I am stuck here. been on the problem for over an hour cant see what to do next. here is my work thus far.
R Ryan Rigdon Junior Member Joined Jun 10, 2010 Messages 246 Sep 20, 2010 #2 i tried it and this is what i came up with. hope its right.
D Deleted member 4993 Guest Sep 20, 2010 #3 \(\displaystyle \int e^u\sin(u)du \ = \ -e^u\cdot cos(u) \ + \ \int e^u\cos(u)du\) \(\displaystyle \int e^u\sin(u)du \ = \ -e^u\cdot cos(u) \ + [e^u\cdot sin(u)\ - \ \int e^u\sin(u)du\) \(\displaystyle \int e^u\sin(u)du \ = \ \frac{1}{2}\cdot \left[\ -e^u\cdot cos(u) \ + e^u\cdot sin(u)\right ]\)
\(\displaystyle \int e^u\sin(u)du \ = \ -e^u\cdot cos(u) \ + \ \int e^u\cos(u)du\) \(\displaystyle \int e^u\sin(u)du \ = \ -e^u\cdot cos(u) \ + [e^u\cdot sin(u)\ - \ \int e^u\sin(u)du\) \(\displaystyle \int e^u\sin(u)du \ = \ \frac{1}{2}\cdot \left[\ -e^u\cdot cos(u) \ + e^u\cdot sin(u)\right ]\)
R Ryan Rigdon Junior Member Joined Jun 10, 2010 Messages 246 Sep 20, 2010 #4 thanx for the hint Subhotosh Khan