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Question Number 185799 by Rupesh123 last updated on 27/Jan/23

Commented by SEKRET last updated on 28/Jan/23

$$\left[\mathbf{x}\right]\to \mathbf{floor}\left(\mathbf{x}\right)?$$$$[1.6]=1$$$$....$$

Answered by MJS_new last updated on 27/Jan/23

$$\underset{0}{\stackrel{2\pi }{\int }}\frac{\sin t}{2+\sin t}dt=$$$$=2\underset{-\pi /2}{\stackrel{\pi /2}{\int }}\frac{\sin t}{2+\sin t}dt$$$$$$$$\int \frac{\sin t}{2+\sin t}dt=$$$$[u=\tan \frac{t}{2}\to dt=\frac{2du}{u^2+1}]$$$$=2\int \frac{u}{(u^2+1)(u^2+u+1)}du=$$$$=2\int \frac{du}{u^2+1}-2\int \frac{du}{u^2+u+1}=$$$$=2\arctan u-\frac{4}{\sqrt{3}}\arctan \frac{2u+1}{\sqrt{4}}$$$$$$$$=2{[2\arctan u-\frac{4}{\sqrt{3}}\arctan \frac{2u+1}{\sqrt{4}}]}_{-1}^1=$$$$=\frac{2(3-2\sqrt{3})\pi }{3}$$

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