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Math Help - Trigonometric integral....

  1. #1
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    Trigonometric integral....

    any help with this?

    \int^{\pi/3}_{0} \frac{\sin^3x}{\sqrt{\cos x}} \, dx

    I first divided the \sin^3 x into two: \frac{\sin^2 x \sin x}{ \sqrt{\cos x}}

    then got \frac{\sin^2 x \sin x}{ \cos^{1/2} x}

    I get confused here but I think it is....

    (-cos^2x) sinx cos^1/2x

    i get stuck here....
    Last edited by mr fantastic; August 24th 2009 at 03:32 AM. Reason: Fixed the latex integral and reformatted other things (that I could make sense of) using latex
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  2. #2
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    Quote Originally Posted by dude487 View Post
    any help with this?

    <br />
\int\frac{sin^3x}{cosx}<br /> <br /> <br />
    Hello,

    I fixed the latex. should be { } not ( )

    have you tried substituting t=\sin(x) ?
    dt=\cos(x) ~dx

    so we have \int\frac{t^3}{\cos^2(x)} ~dt
    and \cos^2(x)=1-\sin^2(x)=1-t^2

    which will give a partial decomposition...


    so let's try u=\cos(x)
    du=-\sin(x) ~dx

    so we have \int \frac{-\sin^2(x)}{u} ~du=\int\frac{u^2-1}{u} ~du=\int u-\frac 1u ~du

    much better ! can you finish it ?


    --------------------------------------------------------
    okay, you changed the problem... however, it doesn't change the method. the best substitution seems to be u=cos(x)
    as for the boundaries, i'll leave it to you =)
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  3. #3
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    Quote Originally Posted by dude487 View Post
    any help with this?

    \int^{\pi/3}_{0} \frac{\sin^3x}{\sqrt{\cos x}} \, dx

    I first divided the \sin^3 x into two: \frac{\sin^2 x \sin x}{ \sqrt{\cos x}}

    then got \frac{\sin^2 x \sin x}{ \cos^{1/2} x}

    I get confused here but I think it is....

    (-cos^2x) sinx cos^1/2x

    i get stuck here....
    \int_0^{\pi /3} {\frac{{{{\sin }^3}x}}{{\sqrt {\cos x} }}dx}  = \int_0^{\pi /3} {\frac{{\sin x{{\sin }^2}x}}{{\sqrt {\cos x} }}dx}  = \int_0^{\pi /3} {\frac{{\sin x\left( {1 - {{\cos }^2}x} \right)}}{{\sqrt {\cos x} }}dx}  =

    = \int_0^{\pi /3} {\sin x\left( {\frac{1}{{\sqrt {\cos x} }} - \sqrt {{{\cos }^3}x} } \right)dx}  = \int_0^{\pi /3} {\left( {\sqrt {{{\cos }^3}x}  - \frac{1}{{\sqrt {\cos x} }}} \right)d\left( {\cos x} \right)}  =

    = \left. {\left( {\frac{2}{5}\sqrt {{{\cos }^5}x}  - 2\sqrt {\cos x} } \right)} \right|_0^{\pi /3} = \left. {2\sqrt {\cos x} \left( {\frac{1}<br />
{5}{{\cos }^2}x - 1} \right)} \right|_0^{\pi /3} =

    = 2 \cdot \frac{1}{{\sqrt 2 }}\left( {\frac{1}{5} \cdot \frac{1}{4} - 1} \right) - 2\left( {\frac{1}{5} - 1} \right) = \sqrt 2 \left( {\frac{1}<br />
{{20}} - 1} \right) - 2\left( {\frac{1}{5} - 1} \right) = \frac{8}<br />
{5} - \frac{{19\sqrt 2 }}{{20}}.
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