My Math Forum INT[min(sinx, cosx)dx] from 0 to pi equals to ??

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 May 1st, 2009, 07:49 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, 05:38 PM #2 Global Moderator   Joined: May 2007 Posts: 6,852 Thanks: 743 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
May 2nd, 2009, 02:32 AM   #3
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Re: INT[min(sinx, cosx)dx] from 0 to pi equals to ??

Quote:
 Originally Posted by mathman min(sinx,cosx) simply means take the smaller one for a given x. In the range 0 to pi, sin is smaller 0

Quote:
 Originally Posted by knowledgegain Thank you very much for your time and help, mathman. I am able to solve the first integral. But for the 2nd integral, i think it could be solved by proper trigonometric substitution, because upper limit is given as 4 and the terms sinx and cosx are there in the integral.

 Tags cosxdx, equals, intminsinx

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# integration of sinxcosx/1 sinxcosx integrate from 0 to pi/2

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