dx-2-sin-x-

Question Number 61719 by aliesam last updated on 06/Jun/19

$$\int\frac{{dx}}{\mathrm{2}+{sin}\left({x}\right)} \\ $$

Commented by maxmathsup by imad last updated on 07/Jun/19

let A = ∫ (dx/(2+sinx)) ⇒A =_(tan((x/2))=t) ∫ (1/(2+((2t)/(1+t^2 )))) ((2dt)/(1+t^2 )) =2 ∫ (dt/(2+2t^2 +2t)) = ∫ (dt/(t^2 +t+1)) =∫ (dt/((t+(1/2))^2 +(3/4))) =_(t+(1/2)=((√3)/2) u) =(4/3)∫ (1/(1+u^2 )) ((√3)/2) du =(2/( (√3))) arctan(u)+c =(2/( (√3))) arctan(((2t+1)/( (√3)))) +c =(2/( (√3))) arctan((2/( (√3)))tan((x/2))+(1/( (√3)))) +c .

$${let}\:{A}\:=\:\int\:\:\frac{{dx}}{\mathrm{2}+{sinx}}\:\Rightarrow{A}\:=_{{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}} \:\:\:\int\:\:\frac{\mathrm{1}}{\mathrm{2}+\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }}\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$=\mathrm{2}\:\int\:\:\:\:\frac{{dt}}{\mathrm{2}+\mathrm{2}{t}^{\mathrm{2}} \:+\mathrm{2}{t}}\:=\:\int\:\:\:\:\frac{{dt}}{{t}^{\mathrm{2}} \:+{t}+\mathrm{1}}\:=\int\:\:\:\:\frac{{dt}}{\left({t}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\frac{\mathrm{3}}{\mathrm{4}}}\:=_{{t}+\frac{\mathrm{1}}{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{u}} \\ $$$$=\frac{\mathrm{4}}{\mathrm{3}}\int\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{1}+{u}^{\mathrm{2}} }\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{du}\:=\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\:{arctan}\left({u}\right)+{c}\:=\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\:{arctan}\left(\frac{\mathrm{2}{t}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)\:+{c} \\ $$$$=\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\:{arctan}\left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}{tan}\left(\frac{{x}}{\mathrm{2}}\right)+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)\:+{c}\:. \\ $$

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