September 9th, 2017, 04:55 AM  #1 
Member Joined: Sep 2013 Posts: 78 Thanks: 2  Integral of sin(x)*cos(x)
Hello I don't understand why I get different results when I try to find that integral. If I choose f' as cos(x) and g as sin(x) then using the partial integration formula the result is (1/2)*sin^2(x). If I choose f' as sin(x) and g as cos(x), then the result is (1/2)*cos^2(x). The two results are not equal, so something is wrong (the 2nd result is correct). Last edited by skipjack; September 9th, 2017 at 12:22 PM. 
September 9th, 2017, 06:06 AM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,939 Thanks: 2266 Math Focus: Mainly analysis and algebra 
Recall that $\sin^2 x + \cos^2 x = 1$ so the two results differ only by an additive constant. Thus they have the same derivative and are therefore both primitives of the same function (your integrand). Note that an easier way to get a (third different) result is to use $\sin 2x = 2\sin x \cos x$ to simplify the integrand before you integrate. 
September 9th, 2017, 06:45 AM  #3 
Member Joined: Sep 2013 Posts: 78 Thanks: 2 
oh,I see,thanks


Tags 
integral, sinxcosx 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Double integral, repeated integral and the FTC  Jhenrique  Calculus  5  June 30th, 2015 03:45 PM 
a new discovery in integral calculus??for integral pros only  gen_shao  Calculus  2  July 31st, 2013 09:54 PM 
integral of double integral in a region E  maximus101  Calculus  0  March 4th, 2011 01:31 AM 
Prove Lower Integral <= 0 <= Upper Integral  xsw001  Real Analysis  1  October 29th, 2010 07:27 PM 
integral of double integral in a region E  maximus101  Algebra  0  December 31st, 1969 04:00 PM 