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May 1st, 2009, 06:49 AM  #1 
Newbie Joined: Apr 2009 Posts: 28 Thanks: 0  INT[min(sinx, cosx)dx] from 0 to pi equals to ??
Dear Members, Any clue how to solve the following integrals? Thanks in advance!! 1. INT[min(sinx, cosx)dx] from 0 to pi equals to (A) 1 ? 2squrt(2) (B) 1 (C) 0 (D) 1 ? squrt(2). [color=#FF40FF]What’s the meaning of min(sinx, cosx), because no other criteria is given?[/color] 2. INT[sinxcosx/(x+1)^2 dx] from 0 to 4 == ? Tried to solve the above integral by 'Integration by Parts'. Since (x+1)^2 in the denominator it's power is increasing. 
May 1st, 2009, 04:38 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,787 Thanks: 708  Re: INT[min(sinx, cosx)dx] from 0 to pi equals to ??
min(sinx,cosx) simply means take the smaller one for a given x. In the range 0 to pi, sin is smaller 0<x<pi/4, while cos is smaller pi/4 < x < pi. You just integrate each part separately and add the two integrals. As for the second integral, you can integrate by parts to go the other way, but it looks somewhat formidable. 
May 2nd, 2009, 01:32 AM  #3  
Newbie Joined: Apr 2009 Posts: 28 Thanks: 0  Re: INT[min(sinx, cosx)dx] from 0 to pi equals to ?? Quote:
Quote:
 

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cosxdx, equals, intminsinx 
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