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Thread: more trigonomentry

  1. #1
    Junior Member TriForce's Avatar
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    more trigonomentry

    $$\frac{tan^3(x)+tan(x)}{tan^3(x)+3tan^2(x)+2tan(x )+6}dx$$

    Help section in the book tells to use substitution t=tan(x) so $$dx=cos^2(t) dt$$:
    $$\frac{t^3+t}{t^3+3t^2+2t+6}cos^2(t) dt$$

    How do i continue from here? Is this a good path to take?
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  2. #2
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    Re: more trigonomentry

    $$d x =\cos ^2x d t$$
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  3. #3
    Junior Member TriForce's Avatar
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    Re: more trigonomentry

    Ah yes, i have trouble with those.

    I can't use $$cos^2(x)$$ right? Coz it's x and im trying to use t. How do i replace dx then?
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  4. #4
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    Re: more trigonomentry

    Quote Originally Posted by TriForce View Post
    Ah yes, i have trouble with those.

    I can't use $$cos^2(x)$$ right? Coz it's x and im trying to use t. How do i replace dx then?
    $dx = \cos^2 x dt$

    If $t = \tan x = \dfrac{\text{opp}}{\text{adj}}$, then $\cos x = \dfrac{\text{adj}}{\text{hyp}} = \dfrac{1}{\sqrt{t^2+1}}$.

    So, $dx = \dfrac{dt}{1+t^2}$
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  5. #5
    Junior Member TriForce's Avatar
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    Re: more trigonomentry

    Solved it, thanks
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