Question Number 185799 by Rupesh123 last updated on 27/Jan/23
Commented by SEKRET last updated on 28/Jan/23
![[x] →floor(x) ? [1.6]= 1 ....](Q185857.png)
$$\left[\boldsymbol{\mathrm{x}}\right]\:\:\rightarrow\boldsymbol{\mathrm{floor}}\left(\boldsymbol{\mathrm{x}}\right)\:\:\:\:? \\ $$$$\:\left[\mathrm{1}.\mathrm{6}\right]=\:\mathrm{1}\:\:\: \\ $$$$.... \\ $$
Answered by MJS_new last updated on 27/Jan/23
![∫_0 ^(2π) ((sin t)/(2+sin t))dt= =2∫_(−π/2) ^(π/2) ((sin t)/(2+sin t))dt ∫((sin t)/(2+sin t))dt= [u=tan (t/2) → dt=((2du)/(u^2 +1))] =2∫(u/((u^2 +1)(u^2 +u+1)))du= =2∫(du/(u^2 +1))−2∫(du/(u^2 +u+1))= =2arctan u −(4/( (√3)))arctan ((2u+1)/( (√4))) =2[2arctan u −(4/( (√3)))arctan ((2u+1)/( (√4)))]_(−1) ^1 = =((2(3−2(√3))π)/3)](Q185802.png)
$$\underset{\mathrm{0}} {\overset{\mathrm{2}\pi} {\int}}\frac{\mathrm{sin}\:{t}}{\mathrm{2}+\mathrm{sin}\:{t}}{dt}= \\ $$$$=\mathrm{2}\underset{−\pi/\mathrm{2}} {\overset{\pi/\mathrm{2}} {\int}}\:\frac{\mathrm{sin}\:{t}}{\mathrm{2}+\mathrm{sin}\:{t}}{dt} \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\int\frac{\mathrm{sin}\:{t}}{\mathrm{2}+\mathrm{sin}\:{t}}{dt}= \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[{u}=\mathrm{tan}\:\frac{{t}}{\mathrm{2}}\:\rightarrow\:{dt}=\frac{\mathrm{2}{du}}{{u}^{\mathrm{2}} +\mathrm{1}}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:=\mathrm{2}\int\frac{{u}}{\left({u}^{\mathrm{2}} +\mathrm{1}\right)\left({u}^{\mathrm{2}} +{u}+\mathrm{1}\right)}{du}= \\ $$$$\:\:\:\:\:\:\:\:\:\:=\mathrm{2}\int\frac{{du}}{{u}^{\mathrm{2}} +\mathrm{1}}−\mathrm{2}\int\frac{{du}}{{u}^{\mathrm{2}} +{u}+\mathrm{1}}= \\ $$$$\:\:\:\:\:\:\:\:\:\:=\mathrm{2arctan}\:{u}\:−\frac{\mathrm{4}}{\:\sqrt{\mathrm{3}}}\mathrm{arctan}\:\frac{\mathrm{2}{u}+\mathrm{1}}{\:\sqrt{\mathrm{4}}} \\ $$$$ \\ $$$$=\mathrm{2}\left[\mathrm{2arctan}\:{u}\:−\frac{\mathrm{4}}{\:\sqrt{\mathrm{3}}}\mathrm{arctan}\:\frac{\mathrm{2}{u}+\mathrm{1}}{\:\sqrt{\mathrm{4}}}\right]_{−\mathrm{1}} ^{\mathrm{1}} = \\ $$$$=\frac{\mathrm{2}\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{3}}\right)\pi}{\mathrm{3}} \\ $$