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March 14th, 2014, 05:44 AM   #1
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Integration by parts

i don't know how to solve the red part
Can somebody teach me please?
Thanks!
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Zynoakib is offline  
 
March 14th, 2014, 06:04 AM   #2
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Re: Integration by parts

Your problem is to integrate . What you give is NOT "integration by parts". You seem to have substituted u for sin(3x) but then left the other "x" without substituting. You can't do that- to integrate you have to have the integral in terms of a single variable.

Integration by parts:
IF you let u= sin(3x), dv= x dx then du= 3cos(x)dx and so


BUT I can see no good reason to do that, since that last integral is harder than the first!

Instead, let u= x, dv= sin(3x)dx so that du= dx, v= -(1/3) cos(3x) and the integral becomes
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March 14th, 2014, 05:21 PM   #3
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I think the procedure by Zynoakib is not necessarily correct.
du = 3 cos(3x) dx and dv = x dx i.e. v = x/2.
I got xsin(3x)/2 - ? x/2 3 cos(3x) dx = xsin(3x)/2 - 3/2 ? x cos(3x) dx.
HallsofIvy correctly pointed out the correct procedure for the integration by parts.
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