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Math Help - Basic questions about integration by substitution

  1. #1
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    Basic questions about integration by substitution

    1. How does it possible for sin(x)cos(x) dx to have two different solutions, namely (1/2)*(sin2x)+c and -(1/2)*(cos2x)+c, so that would bring to (rediculous) conclusion that sin2x+cos2x=0?
    2. Integration by substitution makes use of differentials (not simply derivatives), thus is it purely anti-derivative?
    Thank you in advance
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  2. #2
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    Re: Basic questions about integration by substitution

    Quote Originally Posted by jojo7777777 View Post
    1. How does it possible for sin(x)cos(x) dx to have two different solutions, namely (1/2)*(sin2x)+c and -(1/2)*(cos2x)+c, so that would bring to (rediculous) conclusion that sin2x+cos2x=0?
    You are assuming that the 2 "c"s are the same. Yes, \int sin(x)cos(x) dx= (1/2)sin^2(x)+ c, for some constant, c. Yes, \int sin(x)cos(x)dx= -(1/2)cos^2(x)+ C where "c" and "C" are generally NOT the same. Now that reduces to sin^2(x)+ cos^2(x)=c- C which tells us that sin^2(x)+ cos^2(x) is a constant- NOT "ridiculous".

    2. Integration by substitution makes use of differentials (not simply derivatives), thus is it purely anti-derivative?
    Thank you in advance
    I don't know what you mean by"purely anti-derivative". It is, of course, an anti-derivateve. What does the "purely" indicate? And integration by substitution makes use of both differentials and derivatives which have a "one-to-one correpondence" so it really doesn't matter.
    Last edited by HallsofIvy; February 3rd 2013 at 11:24 AM.
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    Re: Basic questions about integration by substitution

    Thank you very much...!!!
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    Re: Basic questions about integration by substitution

    Moreover 1/2 ( sin^2X) + C = 1/2 ( 1 - cos^2x ) + C = 1/2 -1/2 ( cos^2x) + C = -1/2 (cos^2x) + C +1/2 = -1/2 ( cos^2x) + c where c = C + 1/2
    Thus both the solutions are the same.
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