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Thread: help with integral

  1. #1
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    help with integral

    Hey,

    I was going through my textbook, and came across this challenge question:

    Solve \int{\cos^2{x}dx} without using the half-angle formula.

    I've been running through every trig identity I know, trying to figure out a way to simplify this. Does anyone have an idea? It's trivial to solve using the half-angle formula. I've been racking my brain and can't figure out a way to solve the question without it.
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  2. #2
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    What about \cos^2{x} = \frac{1}{2} + \frac{1}{2}\cos{2x}? The double angle formula...
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    Quote Originally Posted by centenial View Post
    Hey,

    I was going through my textbook, and came across this challenge question:

    Solve \int{\cos^2{x}dx} without using the half-angle formula.

    I've been running through every trig identity I know, trying to figure out a way to simplify this. Does anyone have an idea? It's trivial to solve using the half-angle formula. I've been racking my brain and can't figure out a way to solve the question without it.
    Integration by parts.
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  4. #4
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    Either method will work. I suspect you will find using cos^2(x)= (1/2)(1+ cos(2x)) easier.
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  5. #5
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    Thanks! I guess I was confusing the double and half/angle formulas in my head. That makes sense, pretty straightforward to solve using the double angle.

    mr. fantastic, is there a way to solve that using only integration by parts (without the double angle formula)? If so, I'd be very curious to see how to do that. Cos/sin don't get any simpler when you integrate/differentiate them, so I'm curious to see how that would work.

    Thanks again guys.
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  6. #6
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    Quote Originally Posted by centenial View Post
    Thanks! I guess I was confusing the double and half/angle formulas in my head. That makes sense, pretty straightforward to solve using the double angle.

    mr. fantastic, is there a way to solve that using only integration by parts (without the double angle formula)? If so, I'd be very curious to see how to do that. Cos/sin don't get any simpler when you integrate/differentiate them, so I'm curious to see how that would work.

    Thanks again guys.
    First we'll define our question as being I: I = \int \cos^2(x)\,dx<br />
(this is important later)

    u = \cos(x) \: \rightarrow \: u' = -\sin(x)

    v' = \cos(x) \: \rightarrow \: v = \sin(x)

    Integrating by parts: \int u\,dv = uv - \int v\,du

    I = \sin(x) \cos(x) + \int \sin(x) \sin(x)


    That second integral means you have to integrate by parts again

    p = \sin(x) \: \rightarrow \: p' = \cos(x)

    q' = \sin(x) \: \rightarrow \: q = -\cos(x)

    Thus:

    -\sin(x) \cos(x) + \int \cos^2(x)



    Of course, we earlier defined I = \int cos^2(x) - so substitute it in to remove the integral sign

    I = \sin(x) \cos(x) - (-\sin(x) \cos(x) + I)

    I'll leave you to finish that up, all it needs is algebraic manipulation
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  7. #7
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    Quote Originally Posted by e^(i*pi) View Post
    That second integral means you have to integrate by parts again
    Slightly easier actually to apply pythag at this point. Just in case a picture helps...



    ... where (key in spoiler) ...

    Spoiler:


    ... is the product rule. Straight continuous lines differentiate downwards (integrate up) with respect to x. And,



    ... is lazy integration by parts, doing without u and v.



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