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Thread: trigonometry 3

  1. #1
    Junior Member TriForce's Avatar
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    trigonometry 3

    $$\int\frac{1}{sin^2(x)+2cos^2(x)}dx$$
    My attempt
    $$=\int\frac{1}{2-sin^2(x)}dx = \int\frac{1}{1+cos^2(x)}dx $$
    Not sure how to proceed here so tried substitution t=tan(x/2).
    $$t=tan(x); cos(x)=\frac{1-t^2}{1+t^2}; dx=\frac{2}{1+t^2}dt$$
    But it didn't go well... It seems that there should be a better way. It's also possible that i screwed up.

    $$=\int\frac{1}{1+(\frac{1-t^2}{1+t^2})^2}*\frac{2}{1+t^2}dt$$
    $$=\int\frac{1+t^2}{1+t^4}dt$$

    Does it look ok so far?
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  2. #2
    MHF Contributor
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    Re: trigonometry 3

    No, that does not look correct so far.

    $t = \tan x$, $dx = \dfrac{1}{1+t^2}dt$?

    $\cos^2 x = \dfrac{1}{1+t^2}$

    Then, you have:

    $\displaystyle \int \dfrac{1}{1+ \tfrac{1}{1+t^2} }\cdot \dfrac{dt}{1+t^2} = \int \dfrac{dt}{2+t^2} = \dfrac{1}{\sqrt{2}}\tan^{-1}\left( \dfrac{t}{\sqrt{2}} \right) + C$
    Thanks from topsquark and TriForce
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  3. #3
    Junior Member TriForce's Avatar
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    Re: trigonometry 3

    Thanks I've learned a lot, especially when you explained how cos^2(x) relates to tan(x) and t in another thread here.

    It's interesting how you prefer using t=tanx but our book insist on t=tan(x/2). Says it a general method of integrating all rational functions that have sin and cosx.
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