# Solution to differential equation

I have a differential equation $$x'=\sin(x).$$ WolframAlpha displays the solution as $$2\cot^{-1}(e^{{c_1}-t}),$$ where $c_1$ should be a constant. However, I cannot derive it. I would appreciate any help.

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## 2 Answers

Let $u=e^{c_1-t}$. Then you may verify that, for $x=2\cot^{-1}(u)$, we have $x'(t)=2u/(u^2+1)=\sin(x)$ for whatever $c_1$. So $c_1$ is an undetermined coefficient. This is just like solving $\frac{dy}{dx}=1$, where the solution is $y=x+C$ for some undetermined coefficient $C$.

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$\frac{dx}{dt} = \sin x \rightarrow dt = \frac{dx}{\sin x}$

$t=\int dx \sin^{-1} x = \int dx \sin x \sin^{-2} x = \int \frac{dx \sin x}{1-\cos^2 x}=\int \frac{-du}{1-u^2}$

etc.

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