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Thread: Primitives to trigonometric expression

  1. #1
    Junior Member TriForce's Avatar
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    Primitives to trigonometric expression

    By using substitution t=tan(x/2), find all primitives to:
    $$\frac{1}{(sin(x))^3}$$
    My attempt:
    $$dx=\frac{2}{t^2+1}$$
    $$sin(x)=\frac{2t}{1+t^2}$$
    $$\int\frac{1}{(\frac{2t}{1+t^2})^3}*\frac{2}{t^2+ 1}dt$$
    $$=\frac{1}{4}\int\frac{t^4+2t^2+1}{t^3}dt$$
    After polynomial devision
    $$\frac{1}{4}(\int t + \int\frac{2t^2+1}{t^3}) = \frac{t^2}{8}+\frac{\ln(t)}{2}-\frac{1}{8t^2}$$

    The right anwer is:
    $$-\frac{cosx}{2(sinx)^2}+\frac{1}{2}\ln{(tan(x/2))}+C$$
    Last edited by TriForce; Mar 12th 2018 at 02:24 AM.
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  2. #2
    MHF Contributor
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    Re: Primitives to trigonometric expression

    You haven't attempted to convert your $t$s baxk into $x$s.
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  3. #3
    Junior Member TriForce's Avatar
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    Re: Primitives to trigonometric expression

    $$=\frac{tan^4(\frac{x}{2})-1}{8tan^2(\frac{x}{2})} + \frac{\ln(tan(\frac{x}{2}))}{2}+C$$ ??
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  4. #4
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    Re: Primitives to trigonometric expression

    $\displaystyle \frac{t^2}{8}-\frac{1}{8t^2}=\frac{t^4-1}{8t^2}=-\frac{(1+t^2)(1-t^2)}{8t^2}$

    We know $\displaystyle \cos{x}=\frac{1-t^2}{1+t^2}$ and $\displaystyle \sin{x}=\frac{2t}{1+t^2}$, so try to collect terms that look like that:

    $\displaystyle =-\frac{1+t^2}{2t}\frac{(1-t^2)}{4t}=-\frac{1+t^2}{2t}\frac{1-t^2}{1+t^2}\frac{1+t^2}{2t}\frac{1}{2}=-\frac{\cos{x}}{2\sin^2{x}}$

    Alternatively, you could plug in the reverse substitutions:

    $\displaystyle \tan{\frac{x}{2}}=\frac{\sin{x}}{1+\cos{x}}$

    $\displaystyle \tan{\frac{x}{2}}=\frac{1-\cos{x}}{\sin{x}}$

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