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Math Help - Indefinite Integral

  1. #1
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    Indefinite Integral

    a)

    solve  \int \frac{\sin{2x}}{ \sin^4{x} + \cos^4{x} } ~dx

    by substituting

    i)  u = \tan^2{x}

    ii)  u = \cos{2x}

    b)

    Can we deduce that  \tan^{-1}({\cos{2x}}) = - \tan^{-1}({\tan^2{x}}) ~~ ?
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  2. #2
    MHF Contributor matheagle's Avatar
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    I would let u=\sin x or u=\cos x.

    If u=\sin x, then du=\cos x dx

    and our integral becomes

     \int {\sin{2x} dx\over \sin^4x + \cos^4x} =\int {2\sin x\cos x dx\over \sin^4x + (1-\sin^2x)^2}

      =\int {2u du\over 2u^4 -2u^2 +1}.

    Next, let w=u^2 giving us \int {dw\over 2w^2 -2w +1}

     =\int {dw\over 2(w^2 -w) +1} =\int {dw\over 2(w-1/2)^2 +1/2}.

    From this you can get the arctangent function.
    Last edited by matheagle; June 12th 2009 at 11:19 PM.
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  3. #3
    Moo
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    Hello,

    1. Substituting u=\tan^2(x) is indeed more thorough.

    du=2 \cdot\frac{1}{\cos^2(x)} \cdot \frac{\sin(x)}{\cos(x)} ~dx

    Thus the integral is :

    \int\frac{2\sin(x)\cos(x)}{\sin^4(x)+\cos^4(x)} \cdot\frac{\cos^3(x)}{2\sin(x)} ~du=\int\frac{\cos^4(x)}{\sin^4(x)+\cos^4(x)} ~du=\int\frac{du}{1+u^2}

    2. Substitute u=\cos(2x)

    du=-2\sin(2x) ~dx

    Thus the integral is :

    \int\frac{\sin(2x)}{\sin^4(x)+\cos^4(x)}\cdot\frac  {1}{-2\sin(2x)} ~du=-\frac 12\int\frac{du}{\sin^4(x)+\cos^4(x)}

    Now you have to transform \sin^4(x)+\cos^4(x) with respect to u.

    \cos(2x)=\begin{cases} 1-2\sin^2(x) \\ 2\cos^2(x)-1 \end{cases} \Rightarrow \begin{cases} \sin^4(x)=\left(\frac{1-\cos(2x)}{2}\right)^2=\frac 14 (1-u)^2 \\ \cos^4(x)=\left(\frac{\cos(2x)+1}{2}\right)^2=\fra  c 14 (1+u)^2 \end{cases}

    Thus \sin^4(x)+\cos^4(x)=\frac 14 (2u^2+2)=\frac 12(u^2+1)

    And the integral is :
    -\int\frac{du}{u^2+1}


    And from all this stuff, you can answer question b) (which is not possible with baldeagle's method).
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  4. #4
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    One thing to add - be careful with your constants of integration. Clearly

    <br />
\tan^{-1}({\cos{2x}}) = - \tan^{-1}({\tan^2{x}})

    can't be right - it's not true for x = 0.
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