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Calculus II - Integration by parts

So one of the rules is: ∫f'g dx = f.g - ∫fg'

The prof. did an example question on the board and it was:
∫x.cosx dx
For f', it is cosx, which equals to sinx when integrated. and g=x.
So now, ∫x.cosx dx = sinx.x-∫sinx(x)' dx
The next step, is: =xsinx + cosx + c
Then I completely don't understand this step:
(xsinx +cosx +c)' = sinx +xcosx - sinx?

Can someone explain it to me? Where did the brackets and ' come from? And how did it equal to sinx +xcosx - sinx??
Original post by Airess3

Then I completely don't understand this step:
(xsinx +cosx +c)' = sinx +xcosx - sinx?

Can someone explain it to me? Where did the brackets and ' come from? And how did it equal to sinx +xcosx - sinx??


This just says the differential (indicated by the ') of (xsinx +cosx +c) is sinx +xcosx - sinx.
Original post by SherlockHolmes
This just says the differential (indicated by the ') of (xsinx +cosx +c) is sinx +xcosx - sinx.


Oh ok, now I get it. So the final answer is actually xsinx + cosx + c. And the line after that is just to differentiate it, to check if the final answer is right, because then you end up with xcosx which is what you wanted to integrate at first.
Original post by Airess3
Oh ok, now I get it. So the final answer is actually xsinx + cosx + c. And the line after that is just to differentiate it, to check if the final answer is right, because then you end up with xcosx which is what you wanted to integrate at first.


Yes, exactly.

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