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Problem about integretaion

  1. Jul 17, 2005 #1

    I don't know where is to place this thread. Please can you solve the following integeration problem.
    (Note: sign of integeration is replaced by "S").

    S sin2x/1+sin²x dx

    the answer in the book is "ln(1+sin²x)+c"
  2. jcsd
  3. Jul 17, 2005 #2
    rewrite the numerator using the trig identity sin(2x) = 2 sin(x)cos(x)
    then look for a nice substitution.
  4. Jul 17, 2005 #3
    Could you explain your answer!
  5. Jul 18, 2005 #4

    James R

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    [tex]\int \frac{\sin 2x}{1+sin^2 x}\,dx[/tex]

    Put [itex]u = 1 + sin^2 x[/itex]. Then:

    [tex]du = 2 \sin x \cos x\,dx = \sin 2x\,dx[/tex]

    So, your integral becomes:

    [tex]\int \frac{1}{u}\,du = \ln u + c = \ln (1 + \sin^2 x) + c[/tex]
  6. Jul 18, 2005 #5
    We can do the following:

    [tex] sin(2x) = 2 sin(x)cos(x) [/tex]

    This can be placed in for sin(2x), giving us:

    [tex] \frac{2sin(x)cos(x)}{1 + sin^2(x)} [/tex]

    Now we make a u-substitution:

    [tex] u= 1+ sin^2 (x) [/tex]


    [tex] du = 2sin(x)cos(x)dx [/tex]

    and we can replace the integral as such:

    [tex] \int \frac{1}{u}du [/tex]

    Now this is simply 1/u which equals ln (u)

    so the anwser is ln(u), where u is equal to what we previously stated:

    [tex] ln(1+sin^2(x))+C [/tex]

    where C is the constant of integration.

    I hope that helps you out.



    Edit: Damn, james beat me to it!GRRRRRRRRRRRRRRRRR :rofl:
    Last edited by a moderator: Mar 25, 2008
  7. Jul 18, 2005 #6

    James R

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    Doh! Don't you hate it when that happens! :grumpy:
  8. Jul 18, 2005 #7


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    But it's never a wasted post. Seeing two ways of doing it, even though they're essentially the same, are always better then just one.
  9. Jul 18, 2005 #8

    I agree with you totally on this. I always found it comforting having more than one way to work a problem.
  10. Jul 18, 2005 #9
    Can you solve this too
    S (sinx-cos)² dx

    Answer:- x+sin2x/2 +c
  11. Jul 18, 2005 #10
    Expand it out, and solve the integral, its not hard once you expand it. You will have a sin squared term, plus a cos squared term, which add up to one, plus a -2sinxcosx term. And if you notice, you can use the trig identity from your first question.
  12. Jul 26, 2005 #11
    1) sin2x=2sinxcosx

    2) u=sinx => du=cosx dx

    the intergral come:

    2 * intergral{ u/(1+u^2) du}

    make substitution as again.

    this time let w= 1+u^2 => dw= 2u du

    blah blah....
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