January 24th, 2013, 12:58 AM  #1 
Senior Member Joined: Oct 2012 Posts: 460 Thanks: 0  int x^2sin2x dx
I am given to evaluate: Please, the result is 1/4...I can't get this...should it be just normal multiplying rule for integrals or something else? Many thanks! 
January 24th, 2013, 02:19 AM  #3 
Senior Member Joined: Oct 2012 Posts: 460 Thanks: 0  Re: int x^2sin2x dx
Hello! The problem is: dv= sin2x how do we get v then? so we search what would give us derivative =sin2x, am I right? so sin2x is derivative of v And that is problem! Many thanks! 
January 24th, 2013, 02:25 AM  #4 
Math Team Joined: Apr 2010 Posts: 2,556 Thanks: 296  Re: int x^2sin2x dx
One way is to use [cos(a)]' = sin(a) * [a]'. Sub 2x = a in your equation. Another is use so 
January 24th, 2013, 06:46 AM  #5 
Senior Member Joined: Oct 2012 Posts: 460 Thanks: 0  Re: int x^2sin2x dx
Hello! so: so: [latex]=(\frac{x^2}{2}cosx\int((cosx)[latex] we should get: Can someone help? Many thanks!!! 
January 24th, 2013, 07:57 AM  #6 
Math Team Joined: Apr 2010 Posts: 2,556 Thanks: 296  Re: int x^2sin2x dx
u = x^2 yields du = 2x dx instead of 2. Can you get it from here?

January 24th, 2013, 08:08 AM  #7 
Senior Member Joined: Oct 2012 Posts: 460 Thanks: 0  Re: int x^2sin2x dx
[color=#0000FF]Huh! but the result is up there... not that...please, where is the mistake? [/color] 
January 24th, 2013, 08:28 AM  #8  
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,210 Thanks: 353  Re: int x^2sin2x dx Hello, ungeheuer! Quote:
Integrate by parts: [color=beige]. . [/color] By parts again: [color=beige]. . [/color]  
January 24th, 2013, 09:15 AM  #9 
Senior Member Joined: Oct 2012 Posts: 460 Thanks: 0  Re: int x^2sin2x dx
[b]Many thanks, soroban! All clear now!!! 

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