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Trig Integrals

  1. Aug 31, 2006 #1
    1. [tex] \int^{\frac{\pi}{4}}_{0} sin^{4}x\ cos^{2} x \ dx [/tex]. Would it work to use the half angle formula for both terms? I did this, it took very long. Any quick methods to evaluate this?

    Thanks
     
  2. jcsd
  3. Aug 31, 2006 #2

    quasar987

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    After 5 quick sketchy integrations by parts on a scrap of paper, I get an answer. But I don't guaranty it's not a ghost. You can try if you want. The first step is taking u=sin^4(x)cos(x) dv=cos(x)dx.

    At the end, I get the integral of sin²(x) which is 0.5(x-cosxsinx)
     
  4. Aug 31, 2006 #3

    VietDao29

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    Yes, you can use Power-reduction formulae to evaluate this integral. It'll be a little bit messy, however.
    Or you can do it this way. Here's my approach:
    Let
    [tex]A = \mathop{\int} \limits_{0} ^ {\frac{\pi}{4}} \sin ^ 4 x \cos ^ 2 x dx[/tex]
    And we define another definite integral:
    [tex]B = \mathop{\int} \limits_{0} ^ {\frac{\pi}{4}} \sin ^ 2 x \cos ^ 4 x dx[/tex]
    Now, if we sum the 2 integrals above, we'll have:
    [tex]A + B = \mathop{\int} \limits_{0} ^ {\frac{\pi}{4}} \sin ^ 4 x \cos ^ 2 x dx + \mathop{\int} \limits_{0} ^ {\frac{\pi}{4}} \sin ^ 2 x \cos ^ 4 x dx = \mathop{\int} \limits_{0} ^ {\frac{\pi}{4}} \sin ^ 2 x \cos ^ 2 x (\sin ^ 2 x + \cos ^ 2 x) dx[/tex]
    [tex]= \mathop{\int} \limits_{0} ^ {\frac{\pi}{4}} \sin ^ 2 x \cos ^ 2 x dx = \frac{1}{4} \mathop{\int} \limits_{0} ^ {\frac{\pi}{4}} \sin ^ 2 (2x) dx = ...[/tex]
    And we subtract B from A to get:
    [tex]A - B = \mathop{\int} \limits_{0} ^ {\frac{\pi}{4}} \sin ^ 4 x \cos ^ 2 x dx - \mathop{\int} \limits_{0} ^ {\frac{\pi}{4}} \sin ^ 2 x \cos ^ 4 x dx = \mathop{\int} \limits_{0} ^ {\frac{\pi}{4}} \sin ^ 2 x \cos ^ 2 x (\sin ^ 2 x - \cos ^ 2 x) dx[/tex]
    [tex] = - \mathop{\int} \limits_{0} ^ {\frac{\pi}{4}} \sin ^ 2 x \cos ^ 2 x \cos (2x) dx = - \frac{1}{4} \mathop{\int} \limits_{0} ^ {\frac{\pi}{4}} \sin ^ 2 (2x) \cos (2x) dx = ...[/tex]
    The two integrals above are easier to evaluate than the original one, right?
    Having A + B, and A - B, can you work out A? :)
     
    Last edited: Aug 31, 2006
  5. Aug 31, 2006 #4

    quasar987

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    That was ingenious!
     
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