March 14th, 2014, 05:44 AM  #1 
Newbie Joined: Oct 2013 Posts: 23 Thanks: 0  Integration by parts
i don't know how to solve the red part Can somebody teach me please? Thanks! 
March 14th, 2014, 06:04 AM  #2 
Math Team Joined: Sep 2007 Posts: 2,409 Thanks: 6  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 
March 14th, 2014, 05:21 PM  #3 
Senior Member Joined: Mar 2014 Posts: 112 Thanks: 8 
I think the procedure by Zynoakib is not necessarily correct. du = 3 cos(3x) dx and dv = x dx i.e. v = x²/2. I got x²sin(3x)/2  ? x²/2 3 cos(3x) dx = x²sin(3x)/2  3/2 ? x² cos(3x) dx. HallsofIvy correctly pointed out the correct procedure for the integration by parts. 

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