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anonymous

  • one year ago

What is the integration of cosec2x??

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  1. anonymous
    • one year ago
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    From Mathematica. \[\int\limits \csc (2 x) \, dx=\frac{1}{2} \log (\sin (x))-\frac{1}{2} \log (\cos (x)) \]

  2. IrishBoy123
    • one year ago
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    from standard integral: \(\int \csc{u} \, du = \ln{\left| \csc{u} - \cot{u}\right|} + C\) https://gyazo.com/75a731bf97950a5f69995fa4255a19f7 so for \( \int \csc{2x} \, dx \) sub \(u = 2x, du = 2 dx, dx = du/2\) \(\implies \frac{1}{2}\int \csc{u} \, du = \\\frac{1}{2} \ln{\left| \csc{u} - \cot{u}\right|} + C \\= \frac{1}{2} \ln{\left| \csc{2x} - \cot{2x}\right|} + C\) \(csc{2x} - \cot{2x}\) simplifies \(\large \frac{1}{sin \ 2x} - \frac{cos \ 2x}{sin \ 2x} = \frac{1 - (1- 2 sin^2x)}{2 \ sinx \ cos x} = tan \ x\) \(\implies \int \csc{2x} \, du = \frac{1}{2} \ln{\left| tan \ x\right|} + C\) the sub for the underlying integral follows this original idea for sec https://en.wikipedia.org/wiki/Integral_of_the_secant_function

  3. anonymous
    • one year ago
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    On the off-chance you meant \(\csc^2x\), recall that \(\dfrac{d}{dx}\cot x=-\csc^2x\).

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