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Last edited by 青青子衿 at 2019-6-26 16:52:00\[ f(x)=\int_0^x\frac{\sin t}{3+2\sin t}{\rm\,d}t \]
\[ I(x)=\int\frac{\sin t}{3+2\sin t}{\rm\,d}t \]
\begin{align*}
I(x)&=\tilde{I}(x)+C\\
&=\frac{x}{2}-\frac{3}{\sqrt{5}}\arctan\left(\frac{2+3\tan\frac{x}{2}}{\sqrt{5}}\right)+C\\
\end{align*}
\begin{align*}
f(x)&=\tilde{I}(x)+R(x)\\
&=\frac{x}{2}-\frac{3}{\sqrt{5}}\arctan\left(\frac{2+3\tan\frac{x}{2}}{\sqrt{5}}\right)-\frac{3}{\sqrt{5}}\left(\pi\operatorname{floor}\left(\frac{x}{2\pi}+\frac{1}{2}\right)-\arctan\frac{2}{\sqrt{5}}\right)\\
&=\frac{x}{2}-\frac{3}{\sqrt{5}}\arctan\left(\frac{2+3\tan\frac{x}{2}}{\sqrt{5}}\right)-\frac{3}{\sqrt{5}}\left(\pi\left\lfloor\frac{x}{2\pi}+\frac{1}{2}\right\rfloor-\arctan\frac{2}{\sqrt{5}}\right)
\end{align*}
\begin{align*}
A\cos x+B\sin x
&=\operatorname{sign}\left(A\right)\sqrt{A^2+B^2}\cos\left(x-\arctan\frac{B}{A}\right)\\
&=\operatorname{sign}\left(B\right)\sqrt{A^2+B^2}\sin\left(x+\arctan\frac{A}{B}\right)
\end{align*}
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