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Math Help - how to solve the integration

  1. #1
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    how to solve the integration

    integration of (sinx + cosx)/(sin2x)^(1/2)

    one more sum integration of (sin -1 x)(2)

    Thank u
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: how to solve the integration

    First one:
    I would write it as:
    \frac{1}{\sqrt{2}}\left(\int \frac{\sin(x)}{\sqrt{\sin(x)\cos(x)}}dx+\int \frac{\cos(x)}{\sqrt{\sin(x)\cos(x)}}dx\right)
    =\frac{1}{\sqrt{2}}\left(\int \sqrt{\tan(x)}dx+\int \sqrt{\cot(x)}dx\right)=...
    (but maybe there's an easier way ...)

    Second one:
    If you want to calculate:
    \int \frac{\arcsin(x)}{2}dx
    I would use integration by parts.
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  3. #3
    Ted
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    Re: how to solve the integration

    We have \displaystyle sin(2x)=2tan(x)cos^2(x)

    Which makes the integral be :

    \displaystyle I=\frac{1}{\sqrt{2}} \int \frac{sin(x)+cos(x)}{\sqrt{tan(x)} \, cos(x) } \, dx

    = \displaystyle \frac{1}{\sqrt{2}} \int \frac{tan(x)+1}{\sqrt{tan(x)}} \, dx

    Putting \displaystyle x=arctan(t) gives :

    \displaystyle I=\frac{1}{\sqrt{2}} \int \frac{t+1}{\sqrt{t} (t^2+1)} \, dt

    Putting \displaystyle z=\sqrt{t} gives

    \displaystyle I=\sqrt{2} \int \frac{z^2+1}{z^4+1} \, dz

    We have \displaystyle z^4+1=z^4+2z^2+1-2z^2=(z^2+1)^2-(\sqrt{2} z)^2=(z^2+1-\sqrt{2}z)(z^2+1+\sqrt{2}z)

    Hence,

    \displaystyle I=\int \frac{z^2+1}{(z^2-\sqrt{2}z+1)(z^2+\sqrt{2}z+1)} \, dz

    Use partial fraction.
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  4. #4
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    Re: how to solve the integration

    Thank u for the immediate response. It is very useful. More replies are also invited.
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  5. #5
    MHF Contributor Siron's Avatar
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    Re: how to solve the integration

    Can you show us your steps or results?
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