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Math Help - integral with cotx

  1. #1
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    integral with cotx

    Hi,

    Could someone please give me a hint on how to do this problem? I know that I have to use a formula to do it, but it looks like I need to simplify it.

    Int. cotx/sq. 1+2sinx

    I have these formulas:

    int. cotudu=ln|secu|+c

    int. cot^udu=-1/n-1cot^n-1u-int. cot^n-2udu

    Thank you
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  2. #2
    MHF Contributor red_dog's Avatar
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    \displaystyle\int\frac{\cot x}{\sqrt{1+2\sin x}}dx=\int\frac{\cos x}{\sin x\sqrt{1+2\sin x}}dx
    Substitute \sin x=t\Rightarrow\cos xdx=dt\Rightarrow\int\displaystyle\frac{1}{t\sqrt{  1+2t}}dt.
    Substitute \sqrt{1+2t}=u\Rightarrow 2t=u^2-1\Rightarrow dt=udu
    Then \displaystyle\int\frac{u}{\frac{u^2-1}{2}\cdot u}du=2\int\frac{1}{u^2-1}du=\ln \left|\frac{u-1}{u+1}\right|+C
    Now back substitute.
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  3. #3
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    Hello, chocolatelover!

    I found a method . . . but it took two substitutions.
    I'm certain someone will streamline it for us.


    \int \frac{\cot x}{\sqrt{1+2\sin x}}\,dx

    Let u \,=\,\sin x\quad\Rightarrow\quad x \,=\,\arcsin u\quad\Rightarrow\quad dx \,=\,\frac{du}{\sqrt{1-u^2}}
    . . and: . \cot x \,=\,\frac{\sqrt{1-u^2}}{u}

    Substitute: . \int\frac{\frac{\sqrt{1-u^2}}{u}}{\sqrt{1+2u}}\,\frac{du}{\sqrt{1-u^2}} \;= \;\int\frac{du}{u\sqrt{1+2u}}

    Let v \,=\,\sqrt{1+2u}\quad\Rightarrow\quad u \,=\,\frac{v^2-1}{2}\quad\Rightarrow\quad du \,=\,v\,dv

    Substitute: . \int\frac{v\,dv}{\left(\frac{v^2-1}{2}\right)\cdot v} \;=\;2\int\frac{dv}{v^2-1} \;= \;\ln\left|\frac{v-1}{v+1}\right| + C


    Back-substitute: . \ln\left|\frac{\sqrt{1+2u} - 1}{\sqrt{1+2u} + 1}\right| + C

    Back-substitute: . \boxed{\ln\left|\frac{\sqrt{1+2\sin x} - 1}{\sqrt{1+2\sin x} + 1}\right| + C}


    Ha! . . . red_dog beat me to it!
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  4. #4
    Math Engineering Student
    Krizalid's Avatar
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    Mine just take one.

    Set

    u=\sqrt{1+2\sin x}\implies du=\frac{\cos x}{\sqrt{1+2\sin x}}\,dx, the intregral becomes to

    \int\frac{\cot x}{\sqrt{1+2\sin x}}\,dx=2\int\frac1{u^2-1}\,du

    Cheers,
    K.
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