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Math Help - trig-substitution!

  1. #1
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    trig-substitution!

    i have this  \intk\sqrt{\frac{1-z}{z}}dz

    i set  z=sin^2\theta
    differentiating z w.r.t \theta , i have  dz=2sin\theta cos\theta d\theta
    substituting

     2k\int\sqrt{\frac{1-sin^2\theta}{sin^2\theta}}sin\theta cos\theta d\theta

    i recall that  cos^2\theta=1-sin^2\theta

    i tend to be confuse from here onward, pls help!
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  2. #2
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    Re: trig-substitution!

    you'll get cos^2thetha when you'll put 1-sin^2theta=cos^2theta. then you can expand cos^2theta as (cos2theta+1)/2...then integrate cos2theta/2 and 1/2 separately...
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  3. #3
    Senior Member x3bnm's Avatar
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    Re: trig-substitution!

    \text{You know that } \int \sqrt{\frac{1-z}{z}}dz  = 2\int\sqrt{\frac{1-\sin^{2}(\theta)}{\sin^{2}(\theta)}}\sin(\theta) \cos(\theta)\,\, d\theta

    \text{ by letting } z = \sin^{2}{(\theta)} \text{ and so } dz = 2\sin{(\theta)}\cos{(\theta)}\,\,d\theta

    Now:

    \begin{align*}2\int\sqrt{\frac{1-\sin^{2}(\theta)}{\sin^{2}(\theta)}}\sin(\theta) \cos(\theta)\,\, d\theta  =& 2 \int \frac{\sqrt{(\cos^{2}{(\theta)})}}{\sqrt{(\sin^{2}  {(\theta)})}} \sin(\theta) \cos(\theta)\,\, d\theta..........[\text{ because } 1 - \sin^{2}(\theta) = \cos^{2}(\theta)] \\ =& 2\int\frac{\cos(\theta)}{\sin(\theta)}\sin(\theta) \cos(\theta)\,\, d\theta \\ =& 2\int\cos^{2}(\theta)\,\, d\theta .....[\text{by using simple algebra}] \\ =&  2\int\frac{\cos(2\theta) + 1}{2}\,\, d\theta ......[\text{because }\cos(2\theta)  = 2 \cos^{2}(\theta) - 1]  \\ =& 2\left( \int \frac{\cos(2\theta)}{2}d\theta + \int \frac{1}{2}\,\,d\theta\right) \\ =& 2\left(\frac{\sin(2\theta)}{4} + \frac{\theta}{2}\right) + C \\ =&  \frac{ \sin(2\theta)}{2} + \theta + C \\ =& \frac{2\sin(\theta)\cos(\theta)}{2} + \theta + C ......[\text{because } \sin(2\theta) = 2\sin(\theta)\cos(\theta)] \\ =& \sin(\theta) \cos(\theta) + \theta + C \end{align*}


    By plugging into \sin(\theta) = \sqrt{z}, \cos(\theta) = \sqrt{1 - z} and \theta = \sin^{-1}(\sqrt{z}) we get:

    \int k\sqrt{\frac{1-z}{z}}dz = \sqrt{z} \sqrt{1 - z} + \sin^{-1}(\sqrt{z}) + C

    And that's it. That's the answer. If you differentiate the new found answer with CAS(like maple) you'll get what you started with which is the given integral in the question. Hope this helps.
    Last edited by x3bnm; March 29th 2013 at 08:18 PM.
    Thanks from Ruun
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