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A Integration Problem[Indefinite Integral]

  1. Jan 28, 2013 #1
    1. The problem statement, all variables and given/known data
    Evaluate

    [tex]
    \int (x sin ( x^2 ) )/cos^3 ( x^2 ) dx
    [/tex]



    2. Relevant equations

    none the i know that would be helpful here

    3. The attempt at a solution

    I tried solving this by substitution method by substituting [tex]
    t = x^2[/tex] as t but i dont know how to proceed after that :confused:
     
  2. jcsd
  3. Jan 28, 2013 #2

    LCKurtz

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    Show us what you got, because you should be able to push that through. Alternatively, try ##t=\cos(x^2)##.
     
  4. Jan 28, 2013 #3
    Damn how could i not see that :rolleyes:
    okay i got the answer as ##I= 1/4(1/cos^2(x^2)## :biggrin:

    Can anyone confirm this :uhh:
     
  5. Jan 28, 2013 #4

    LCKurtz

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    You don't need someone else to confirm it. Differentiate your answer and see if it works.
     
  6. Jan 28, 2013 #5

    SammyS

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    Don't forget the constant of integration.
     
  7. Jan 29, 2013 #6

    HallsofIvy

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    If you had gone with your first idea, [itex]t= x^2[/itex] so that dt= 2x dx and (1/2)dt= x dx, you would have got [itex](1/2)\int sin(t)/cos^3(t) dt[/itex]. Now, it should be clear, with that "sin(t)dt" in the numerator, you can let y= cos(t), so that dy= sin(t) dt, and the integral becomes [itex](1/2)\int dy/y^3= (1/2)\int y^{-3}dy[/itex].

    Essentially, you are doing the same thing in two steps rather than one.
     
  8. Jan 29, 2013 #7
    Sir correct me if i am wrong but i think there is a mistake in the method you have done which i first stated.

    if we took [itex]t= x^2[/itex] and dt =2x dx we cannot substitute that in the equation because ##x^2## is within the sine function and dx is outside??? :confused:
     
  9. Jan 29, 2013 #8

    Zondrina

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    Wait wait wait. Listen to what you're saying here. You said t = x2 right? So everywhere x2 appears in your equation you replace it with t.

    Now dt = 2x dx which implies that (1/2)dt = xdx. So anywhere you see xdx ( which it's VERY apparent to see ), you replace it with (1/2)dt.

    Clear now?
     
  10. Jan 29, 2013 #9

    CompuChip

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    You replace all occurrences of x separately.

    So the sine and cosine contain an x², which you substitute by t.

    Then outside the trig functions, you are left with an x dx which - as you said - you substitute by (1/2) dt.
     
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