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Math Help - Very simple trigonometric integral

  1. #1
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    Question Very simple trigonometric integral

    Now you will all point at me and laugh. I tried to prove that \int_{0}^{\pi}\sin^{3}\theta d\theta=\frac{4}{3} and all I get is
    \int_{0}^{\pi}\sin^{3}\theta d\theta=\left[3\sin^{2}\theta\cos\theta\right]_{0}^{\pi}=3*0^{2}*(-1)-3*0^{2}*1=0

    Where does it all go so wrong?
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  2. #2
    Super Member Quacky's Avatar
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    Re: Very simple trigonometric integral

    Quote Originally Posted by fysikbengt View Post
    Now you will all point at me and laugh. I tried to prove that \int_{0}^{\pi}\sin^{3}\theta d\theta=\frac{4}{3} and all I get is
    \int_{0}^{\pi}\sin^{3}\theta d\theta=\left[3\sin^{2}\theta\cos\theta\right]_{0}^{\pi}=3*0^{2}*(-1)-3*0^{2}*1=0

    Where does it all go so wrong?
    You've differentiated; you wanted to integrate. To integrate this, I would either write it as  Sin\theta(1-Cos^2(\theta)) (and expand the brackets) or use some sort of double angle formula to manipulate.
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  3. #3
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    Re: Very simple trigonometric integral

    Quote Originally Posted by fysikbengt View Post
    Now you will all point at me and laugh. I tried to prove that \int_{0}^{\pi}\sin^{3}\theta d\theta=\frac{4}{3} and all I get is
    \int_{0}^{\pi}\sin^{3}\theta d\theta=\left[3\sin^{2}\theta\cos\theta\right]_{0}^{\pi}=3*0^{2}*(-1)-3*0^{2}*1=0

    Where does it all go so wrong?
    Maybe the reason you're getting the wrong answer is because \displaystyle \int{\sin^3{\theta}\,d\theta} \neq 3\sin^2{\theta}\cos{\theta}...
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  4. #4
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    Re: Very simple trigonometric integral

    You can also use the identity \sin(3\theta) = 3\sin(\theta) - 4\sin^3(\theta) which I suspect Quacky was alluding to since \sin(\theta) \text{  and  } \sin(a\theta) (where a is a constant) are simple to integrate.

    You get bonus kudos for deriving that identity yourself
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    Re: Very simple trigonometric integral

    Quote Originally Posted by e^(i*pi) View Post
    You can also use the identity \sin(3\theta) = 3\sin(\theta) - 4\sin^3(\theta) which I suspect Quacky was alluding to since \sin(\theta) \text{  and  } \sin(a\theta) (where a is a constant) are simple to integrate.

    You get bonus kudos for deriving that identity yourself
    I'll use the lazy derivation and leave the proper derivation for the OP

    \displaystyle \begin{align*} (\cos{\theta} + i\sin{\theta})^3 &= \cos^3{\theta} + 3i\cos^2{\theta}\sin{\theta} + 3i^2\cos{\theta}\sin^2{\theta} + i^3\sin^3{\theta} \\ &= \cos^3{\theta} - 3\cos{\theta}\sin^2{\theta} + i(3\cos^2{\theta}\sin{\theta} - \sin^3{\theta}) \\ &= \cos^3{\theta} - 3\cos{\theta}(1 - \cos^2{\theta}) + i[3(1 - \sin^2{\theta})\sin{\theta} - \sin^3{\theta}] \\ &= \cos^3{\theta} - 3\cos{\theta} + 3\cos^3{\theta} + i(3\sin{\theta} - 3\sin^3{\theta} - \sin^3{\theta}) \\ &= 4\cos^3{\theta} - 3\cos{\theta} + i(3\sin{\theta} - 4\sin^3{\theta}) \end{align*}


    But by DeMoivre's Theorem, we also have

    \displaystyle (\cos{\theta} + i\sin{\theta})^3 = \cos{3\theta} + i\sin{3\theta}.


    Equating real and imaginary parts gives

    \displaystyle \cos{3\theta} = 4\cos^3{\theta} - 3\cos{\theta}

    and

    \displaystyle \sin{3\theta} = 3\sin{\theta} - 4\sin^3{\theta}
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    Lightbulb Re: Very simple trigonometric integral

    Quote Originally Posted by e^(i*pi) View Post
    You can also use the identity \sin(3\theta) = 3\sin(\theta) - 4\sin^3(\theta) which I suspect Quacky was alluding to since \sin(\theta) \text{  and  } \sin(a\theta) (where a is a constant) are simple to integrate.

    You get bonus kudos for deriving that identity yourself
    Sure! By looking at the unit circle you can prove that \cos (x+y)=\cos x \cos y -\sin x \sin y and by substituting x with x-\pi /2 this leads to \sin(x+y)=\sin x \cos y +\cos x \sin y. From these two formulae you can easily prove that \sin(2x)=2\sin x\cos x and \cos(2x)=1-2\sin^{2}x.

    Now the rest follows:
    \sin(3x)=\sin(x+2x)=\sin x\cos(2x)+\cos x\sin(2x)=\sin x(1-2\sin^{2}x)+\cos x 2\sin x\cos x=\sin x -2\sin^{3}x+2\sin x (1-\sin^{2}x)=3\sin x -4\sin^{3}x
    And that's it.
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