## anonymous one year ago What is the integration of cosec2x??

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1. sepeario
2. anonymous

From Mathematica. $\int\limits \csc (2 x) \, dx=\frac{1}{2} \log (\sin (x))-\frac{1}{2} \log (\cos (x))$

3. IrishBoy123

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

4. anonymous

On the off-chance you meant $$\csc^2x$$, recall that $$\dfrac{d}{dx}\cot x=-\csc^2x$$.