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Need help on integral problem

  1. Dec 21, 2012 #1
    1. The problem statement, all variables and given/known data
    ∫cos^3xdx


    2. Relevant equations


    3. The attempt at a solution
    I started teaching myself integral calculus yesterday and I had no idea what to do for this problem... I tried splitting cos^3x into cos^2x(cosx) but that didn't work. How do you solve this?
     
  2. jcsd
  3. Dec 21, 2012 #2

    dextercioby

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    You need to split it the way you did it and use trigonometry. [itex] \cos^2 x = 1-\sin^2 x [/itex] plus substitution.
     
  4. Dec 21, 2012 #3
    [itex]\displaystyle \int cos^n {x} \ dx = \frac{sin{x} \ cos^{n-1}{x}}{n} + \frac{n-1}{n} \int cos^{n-2}x \ dx[/itex].

    Alternatively, [itex]\displaystyle \int cos^3 {x} \ dx = \int cos^2 x \ cosx \ dx = \int (1-sin^2 x) cos x \ dx[/itex]. From here, you could use substitution.
     
  5. Dec 21, 2012 #4

    sharks

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    Do you know about the triple angle formulas? Maybe you've come across it in trigonometry:
    $$\cos 3\theta =4\cos^3 \theta - 3\cos \theta$$
    Re-arranging:
    $$cos^3 \theta = \frac{1}{4} (\cos 3\theta + 3\cos \theta)$$
    Therefore,
    $$\int cos^3 \theta\,.d\theta= \frac{1}{4} \int (\cos 3\theta + 3\cos \theta)\,.d\theta$$
     
  6. Dec 21, 2012 #5

    haruspex

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    ... or just use cos x dx = d sin x.
     
  7. Dec 22, 2012 #6

    HallsofIvy

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    Yes, that's what madelbroth meant.
     
  8. Dec 22, 2012 #7

    haruspex

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    I know, just saying you don't have to go through a formal substitution, adjusting limits.
     
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