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Math Help - Integration of 3rd degree trig functions?

  1. #1
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    Integration of 3rd degree trig functions?

    I just got back from my diff eq exam.

    One of the problems was y'+y*(tan(x)) = 8*sin(x^3)

    At a first glance, it looks like a simple first order linear diff eq. But after getting an integrating factor of sec(x), how do I find the integral of 8*sin(x^3)cos(x)?

    Wolfram alpha seems to be returning a cryptic answer.
    http://www.wolframalpha.com/input/?i=y%27%2By*%28tan%28x%29%29+%3D+8*sin%28x^3%29
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    Re: Integration of 3rd degree trig functions?

    Quote Originally Posted by astroidea View Post
    I just got back from my diff eq exam.

    One of the problems was y'+y*(tan(x)) = 8*sin(x^3)

    At a first glance, it looks like a simple first order linear diff eq. But after getting an integrating factor of sec(x), how do I find the integral of 8*sin(x^3)cos(x)?

    Wolfram alpha seems to be returning a cryptic answer.
    http://www.wolframalpha.com/input/?i=y%27%2By*%28tan%28x%29%29+%3D+8*sin%28x^3%29
    Are you sure the right hand side is sin(x^3) and not [sin(x)]^3? If it's the latter, the integral is a simple u substitution...
    Thanks from astroidea
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    Re: Integration of 3rd degree trig functions?

    Pardon my ineptness at math. I had it confused with the sin^2(x) and [sin(x)]^2 identity. But yes, it's [sin(x)]^3

    How do you do the U substitution? With sin^3(x)dx/cos(x) let cos(x)=u, -du= sinxdx, this doesn't seem to work since there will still be sin^2x left

    Let u = sin(x), du=cosxdx... I can't get that to work since it's dx/cos(x).

    MUCHO thanks for the help
    Last edited by astroidea; March 30th 2012 at 02:40 PM.
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    Re: Integration of 3rd degree trig functions?

    Quote Originally Posted by astroidea View Post
    Pardon my ineptness at math. I had it confused with the sin^2(x) and [sin(x)]^2 identity. But yes, it's [sin(x)]^3

    How do you do the U substitution? With sin^3(x)dx/cos(x) let cos(x)=u, -du= sinxdx, this doesn't seem to work since there will still be sin^2x left

    Let u = sin(x), du=cosxdx... I can't get that to work since it's dx/cos(x).

    MUCHO thanks for the help
    \displaystyle \begin{align*} \frac{dy}{dx} + y\tan{x} = 8\sin^3{x} \end{align*}

    The integrating factor is \displaystyle \begin{align*} e^{\int{\tan{x}\,dx}} = e^{\ln{(\sec{x})}} = \sec{x} \end{align*}, so

    \displaystyle \begin{align*} \sec{x}\,\frac{dy}{dx} + y\sec{x}\tan{x} &= 8\sin^3{x}\sec{x} \\ \frac{d}{dx}\left(y\sec{x}\right) &= 8\sin^3{x}\sec{x} \\ y\sec{x} &= \int{8\sin^3{x}\sec{x}\,dx} \\ y\sec{x} &= 8\int{\frac{\sin{x}\left(1 - \cos^2{x}\right)}{\cos{x}}\,dx} \\ y\sec{x} &= 8\int{\frac{\sin{x} - \sin{x}\cos^2{x}}{\cos{x}}\,dx} \\ y\sec{x} &= 8\int{\frac{\sin{x}}{\cos{x}} - \frac{\sin{x}\cos^2{x}}{\cos{x}}\,dx } \\ y\sec{x} &= 8\int{\frac{\sin{x}}{\cos{x}} - \sin{x}\cos{x}\,dx } \end{align*}

    You can now solve each of those with a u substitution
    Thanks from astroidea
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  5. #5
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    Re: Integration of 3rd degree trig functions?

    Ah, I must brush up on my trig identities.
    Thanks a lot for demystifying this problem for me!
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