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Math Help - Integration of Trigonometric Identities - Yay!

  1. #1
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    Integration of Trigonometric Identities - Yay!

    Remind why I'm taking further maths next year >.<

    Question time!

    I'm currently doing, as you probably noticed in the title (:P) integration of trig identities. I've come across this lovely little question which asks me to:

    Find the integral of cos2xcosx

    (I have no idea how to stick the integral sign in a text page ^_^)

    Would somebody be so kind as to explain to me how to do this?

    Many thanks
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  2. #2
    Moo
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    Hello,

    Note that cos(2x)=1-2sinē(x)

    Hence cos(2x)cos(x)=cos(x)-2cos(x)sinē(x)

    You know the primitive of cos(x) (at least i hope ^^)

    For the second term, cos is the derivative of sin, so you have something of the form :

    k \ u'(x) \ u^2(x), which gives a primitive : k \frac13 u^3(x)

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  3. #3
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    I'm just going to write down my chain of thought - could you please correct me if I go awry

    cos(x) - 2cos(x)sinē(x)
    cos(x) - 2cos(x)(1 - cosē(x))
    cos(x) - 2cos(x) - 2cos^3 (x)
    -cos(x) - 2cos^3 (x)

    Is this right so far in terms of deriving etc. ?
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  4. #4
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    Never mind, I'm going to call it a night and face the unparalleled wrath of an angry maths teacher

    Thanks for the help guys ^_^ I have enough to make it look like I tried :P

    X
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  5. #5
    Moo
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    Well, guess it's too late ~

    cos(2x)=1-2sinē(x)

    -> cos(2x)cos(x)=cos(x)-2cos(x)sinē(x)

    \int \cos(x) - 2 \cos(x) \sin^2(x) dx = \underbrace{\int \cos(x) dx}_{\sin(x)} - 2 \int \underbrace{\cos(x)}_{u'(x)} \underbrace{\sin^2(x)}_{u^2(x)} dx

    = \sin(x) - 2*3 \int \frac13 u'(x) u^2(x) dx = \sin(x) - 6 [u^3(x)] = \sin(x)-6 \sin^3(x)
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