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Math Help - Calculus on trigonometry... help!

  1. #1
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    Calculus on trigonometry... help!

    "integral"{cos(2x)/[cos^2(x)*sin^2(x)]}*dx
    Full-length solution for this would be really helpful!
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  2. #2
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    Re: Calculus on trigonometry... help!

    Expand cos(2x) to get it in terms of angle x instead of 2x
    Use the substitution u=cos^2x and make use of cos^2x+sin^2x=1
    Thanks from HallsofIvy
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  3. #3
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    Re: Calculus on trigonometry... help!

    u = cos^2(x) or u = cos (x) ?
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  4. #4
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    Re: Calculus on trigonometry... help!

    Hello, yugimutoshung!

     \int \frac{\cos(2x)}{\cos^2(x)\sin^2(x)}\,dx

    Note that: . \cos^2(x)\sin^2(x) \:=\:\big[\sin(x)\cos(x)\big]^2

    Also that: . \sin(x)\cos(x) \:=\:\tfrac{1}{2}\sin(2x)

    The denominator becomes: . \left(\tfrac{1}{2}\sin(2x)\right)^2 \:=\:\tfrac{1}{4}\sin^2(2x)

    The integral becomes: . \int\frac{\cos(2x)\,dx}{\frac{1}{4}\sin^2(2x)} \;=\;4\int\frac {\cos(2x)\,dx}{\sin^2(2x)}

    Let u \,=\,\sin(2x) \quad\Rightarrow\quad du \,=\,2\cos(2x)\,dx \quad\Rightarrow\quad \cos(2x)\,dx \,=\,\tfrac{1}{2}du

    Substitute: . 4\int\frac{\frac{1}{2}du}{u^2} \;=\;2\int u^{-2}du \;=\;2\left(\frac{u^{-1}}{-1}\right) + C \;=\;-\frac{2}{u}+C

    Back-substitute: . -\frac{2}{\sin(2x)} + C \;=\;-2\csc(2x) + C
    Thanks from dokrbb, yugimutoshung and abualabed
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  5. #5
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    Re: Calculus on trigonometry... help!

    Beautifully done, Soroban! Thanks!
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