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Question Number 72060 by Tushar rathore last updated on 23/Oct/19
∫((sin x)/(1+sin x+sin 2x))dx=?
$$\int\frac{\mathrm{sin}\:{x}}{\mathrm{1}+\mathrm{sin}\:{x}+\mathrm{sin}\:\mathrm{2}{x}}{dx}=? \\ $$
Commented by Tushar rathore last updated on 27/Oct/19
anyone tried this
$${anyone}\:{tried}\:{this} \\ $$
Commented by mathmax by abdo last updated on 27/Oct/19
let A =∫((sinx)/(1+sinx +sin(2x)))dx ⇒ A =∫ ((sinx)/(1+sinx +2sinx cosx))dx =_(tan((x/2))=t) ∫ (((2t)/(1+t^2 ))/(1+((2t)/(1+t^2 )) +((2(2t)(1−t^2 ))/((1+t^2 )^2 )))) ((2dt)/(1+t^2 )) = ∫ ((4t)/((1+t^2 )^2 { 1+((2t)/(1+t^2 )) +((4t(1−t^2 ))/((1+t^2 )^2 ))}))dt =4∫ (t/((1+t^2 )^2 +2t(1+t^2 ) +4t(1−t^2 )))dt =4 ∫ ((tdt)/(t^4 +2t^2 +1 +2t+2t^3 +4t−4t^3 )) =4 ∫ ((tdt)/(t^4 −2t^3 +2t^2 +6t +1)) rest decomposition of F(t) =(t/(t^4 −2t^3 +2t^2 +6t +1)) the roots of p(t)=t^4 −2t^3 +2t^2 +6t +1 are t_1 =1,5898 +1,7445i (complex) t_2 =1,5898−1,7445i(complex) t_3 =−1 (real) t_4 =−0,1795(real) ⇒F(t)=(t/((t+1)(t−t_4 )(t−t_1 )(t−t_1 ^− ))) =(t/((t+1)(t−t_4 )(t^2 −2Re(t_1 )t +∣t_1 ∣^2 ))) =(a/(t+1)) +(b/(t−t_4 )) +((ct +d)/(t^2 −2Re(t_1 )t +∣t_1 ∣^2 )) ...be continued...
$${let}\:{A}\:=\int\frac{{sinx}}{\mathrm{1}+{sinx}\:+{sin}\left(\mathrm{2}{x}\right)}{dx}\:\Rightarrow\:{A}\:=\int\:\:\:\frac{{sinx}}{\mathrm{1}+{sinx}\:+\mathrm{2}{sinx}\:{cosx}}{dx} \\ $$$$=_{{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}} \:\:\:\:\:\int\:\:\:\frac{\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }}{\mathrm{1}+\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }\:+\frac{\mathrm{2}\left(\mathrm{2}{t}\right)\left(\mathrm{1}−{t}^{\mathrm{2}} \right)}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }}\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$=\:\int\:\:\:\:\:\frac{\mathrm{4}{t}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} \left\{\:\mathrm{1}+\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }\:+\frac{\mathrm{4}{t}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\right\}}{dt} \\ $$$$=\mathrm{4}\int\:\:\:\:\frac{{t}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} \:+\mathrm{2}{t}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\:+\mathrm{4}{t}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)}{dt} \\ $$$$=\mathrm{4}\:\int\:\:\:\frac{{tdt}}{{t}^{\mathrm{4}} \:+\mathrm{2}{t}^{\mathrm{2}} \:+\mathrm{1}\:+\mathrm{2}{t}+\mathrm{2}{t}^{\mathrm{3}} \:+\mathrm{4}{t}−\mathrm{4}{t}^{\mathrm{3}} } \\ $$$$=\mathrm{4}\:\int\:\:\:\:\frac{{tdt}}{{t}^{\mathrm{4}} −\mathrm{2}{t}^{\mathrm{3}} \:+\mathrm{2}{t}^{\mathrm{2}} \:+\mathrm{6}{t}\:+\mathrm{1}}\:\:{rest}\:{decomposition}\:{of}\: \\ $$$${F}\left({t}\right)\:=\frac{{t}}{{t}^{\mathrm{4}} −\mathrm{2}{t}^{\mathrm{3}} \:+\mathrm{2}{t}^{\mathrm{2}} \:+\mathrm{6}{t}\:+\mathrm{1}}\:{the}\:{roots}\:{of}\:{p}\left({t}\right)={t}^{\mathrm{4}} −\mathrm{2}{t}^{\mathrm{3}} \:+\mathrm{2}{t}^{\mathrm{2}} \:+\mathrm{6}{t}\:+\mathrm{1} \\ $$$${are}\:{t}_{\mathrm{1}} =\mathrm{1},\mathrm{5898}\:+\mathrm{1},\mathrm{7445}{i}\:\left({complex}\right) \\ $$$${t}_{\mathrm{2}} =\mathrm{1},\mathrm{5898}−\mathrm{1},\mathrm{7445}{i}\left({complex}\right) \\ $$$${t}_{\mathrm{3}} =−\mathrm{1}\:\left({real}\right)\: \\ $$$${t}_{\mathrm{4}} =−\mathrm{0},\mathrm{1795}\left({real}\right)\:\Rightarrow{F}\left({t}\right)=\frac{{t}}{\left({t}+\mathrm{1}\right)\left({t}−{t}_{\mathrm{4}} \right)\left({t}−{t}_{\mathrm{1}} \right)\left({t}−\overset{−} {{t}}_{\mathrm{1}} \right)} \\ $$$$=\frac{{t}}{\left({t}+\mathrm{1}\right)\left({t}−{t}_{\mathrm{4}} \right)\left({t}^{\mathrm{2}} −\mathrm{2}{Re}\left({t}_{\mathrm{1}} \right){t}\:+\mid{t}_{\mathrm{1}} \mid^{\mathrm{2}} \right)}\:=\frac{{a}}{{t}+\mathrm{1}}\:+\frac{{b}}{{t}−{t}_{\mathrm{4}} }\:+\frac{{ct}\:+{d}}{{t}^{\mathrm{2}} −\mathrm{2}{Re}\left({t}_{\mathrm{1}} \right){t}\:+\mid{t}_{\mathrm{1}} \mid^{\mathrm{2}} } \\ $$$$…{be}\:{continued}… \\ $$

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