Solved: What Was the Missing Detail in This Integration by Parts Problem?

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The discussion revolves around a problem involving integration by parts, where the user initially felt uncertain about the definition. They identified that the main issue was a missed detail in the integration process, specifically related to the integration variables pi and cero. The user clarified their approach using partial integration, noting that the sine terms ultimately equate to zero. A point of confusion arose regarding the omission of the Omega term when transitioning between steps in the solution. The expected answer for the problem was stated to be 4/pi.
paul-martin
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sorry that i disturbed you guys with this, of course the problem as always was simpel i just missed one little detail, i was at first scared i had missed something in the defintion of integration by parts.

Thx for your time!
 
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It can be hard to se the integration variables which is pi and cero. I am using partial integration there.
 
It is nothing advanced sen all the sins turn out to be cero.
 
I don't understand why you dropped the \Omega when proceeding from Step 4 to Step 5.
 
http://img388.imageshack.us/img388/4140/integration29jv.jpg

Omega=2*pi/T and T= 2pi
 
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I should mention the answer should be 4/pi
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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