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integral of cot(x)

What is the integral of cot(x)?

The integral of cot(x)\cot(x) is lnsin(x)+C\ln \lvert \sin(x) \rvert + C.

Proof. By definition, we have that cot(x)=1tan(x)=cos(x)sin(x)\cot(x) = \frac{1}{\tan(x)} = \frac{\cos(x)}{\sin(x)}. So:
cot(x)dx=1tan(x)dx=cos(x)sin(x)dx.\begin{align*} \int \cot(x) dx = \int \frac{1}{\tan(x)} dx = \int \frac{\cos(x)}{\sin(x)} dx. \end{align*}
Now we want to apply the substitution method. Let u=sin(x)u = \sin(x). Then we get:
ddxu=cos(x)    du=cos(x)dx.\begin{align*} \frac{d}{dx} u = \cos(x) \iff du = \cos(x)dx. \end{align*}
Together, we have that:
cot(x)dx=1tan(x)dx=cos(x)sin(x)dx=1udu=lnu+C=lnsin(x)+Csince u=sin(x).\begin{align*} \int \cot(x) dx = \int \frac{1}{\tan(x)} dx &= \int \frac{\cos(x)}{\sin(x)} dx \\ &= \int \frac{1}{u} du \\ &= \ln \lvert u \rvert + C \\ &= \ln \lvert \sin(x) \rvert + C \quad \text{since } u = \sin(x). \end{align*}
Therefore, the integral of cot(x)\cot(x) is lnsin(x)+C\ln \lvert \sin(x) \rvert + C.

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