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Math Help - Some help with a relatively straight forward integral please!

  1. #1
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    Some help with a relatively straight forward integral please!

    Just trying to understand the answer in my book here...

    given that cos(2x) = 2cos^2 x -1

    then:

    cos^2 x = 1/2(cos(2x) + 1). agreed.

    my problem comes upon integrating cos^2 x..


    i get thus far:


    cos^2 x dx = 1/2(cos(2x) + 1).

    =1/2(1/2 sin(2x) + x) + c

    agreed. it is the next bit that is throwing me off..

    distributing out the 1/2 i get:

    1/4sin(2x) + 2x) + c

    i'm sure i'm missing something terribly obvious here, but where does the the extra x (2x term) pop up from ?

    kind thankyous in advance!
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  2. #2
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    Re: Some help with a relatively straight forward integral please!

    Quote Originally Posted by euphmorning View Post
    Just trying to understand the answer in my book here...

    given that cos(2x) = 2cos^2 x -1

    then:

    cos^2 x = 1/2(cos(2x) + 1). agreed.

    my problem comes upon integrating cos^2 x..


    i get thus far:


    cos^2 x dx = 1/2(cos(2x) + 1).

    =1/2(1/2 sin(2x) + x) + c

    agreed. it is the next bit that is throwing me off..

    distributing out the 1/2 i get:

    1/4sin(2x) + 2x) + c

    i'm sure i'm missing something terribly obvious here, but where does the the extra x (2x term) pop up from ?

    kind thankyous in advance!
    The antiderivative of 1 is x. Also, when you should distribute you should get (1/2)x, not 2x.
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  3. #3
    MHF Contributor Reckoner's Avatar
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    Re: Some help with a relatively straight forward integral please!

    Quote Originally Posted by euphmorning View Post
    1/4sin(2x) + 2x) + c

    i'm sure i'm missing something terribly obvious here, but where does the the extra x (2x term) pop up from ?
    I think you may be missing a parenthesis there. I'm assuming that you're trying to factor out the \textstyle\frac12?

    \frac12\left(\frac12\sin2x + x\right) + C

    =\frac12\left(\frac12\sin2x + \frac12\cdot2x\right) + C

    =\frac12\cdot\frac12\left(\sin2x + 2x\right) + C

    =\frac14\left(\sin2x + 2x\right) + C
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  4. #4
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    Re: Some help with a relatively straight forward integral please!

    [QUOTE=Reckoner;724839]I think you may be missing a parenthesis there. I'm assuming that you're trying to factor out the \textstyle\frac12?

    \frac12\left(\frac12\sin2x + x\right) + C

    =\frac12\left(\frac12\sin2x + \frac12\cdot2x\right) + C

    =\frac12\cdot\frac12\left(\sin2x + 2x\right) + C

    [tex]=\frac14\left(\sin2x + 2x\right) + C

    Thanks alot! That sorted it out nicely
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