Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

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...

$${let}\:{A}\:=\int\frac{{sinx}}{\mathrm{1}+{sinx}\:+{sin}\left(\mathrm{2}{x}\right)}{dx}\:\Rightarrow\:{A}\:=\int\:\:\:\frac{{sinx}}{\mathrm{1}+{sinx}\:+\mathrm{2}{sinx}\:{cosx}}{dx} \\ $$$$=_{{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}} \:\:\:\:\:\int\:\:\:\frac{\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }}{\mathrm{1}+\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }\:+\frac{\mathrm{2}\left(\mathrm{2}{t}\right)\left(\mathrm{1}−{t}^{\mathrm{2}} \right)}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }}\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$=\:\int\:\:\:\:\:\frac{\mathrm{4}{t}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} \left\{\:\mathrm{1}+\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }\:+\frac{\mathrm{4}{t}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\right\}}{dt} \\ $$$$=\mathrm{4}\int\:\:\:\:\frac{{t}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} \:+\mathrm{2}{t}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\:+\mathrm{4}{t}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)}{dt} \\ $$$$=\mathrm{4}\:\int\:\:\:\frac{{tdt}}{{t}^{\mathrm{4}} \:+\mathrm{2}{t}^{\mathrm{2}} \:+\mathrm{1}\:+\mathrm{2}{t}+\mathrm{2}{t}^{\mathrm{3}} \:+\mathrm{4}{t}−\mathrm{4}{t}^{\mathrm{3}} } \\ $$$$=\mathrm{4}\:\int\:\:\:\:\frac{{tdt}}{{t}^{\mathrm{4}} −\mathrm{2}{t}^{\mathrm{3}} \:+\mathrm{2}{t}^{\mathrm{2}} \:+\mathrm{6}{t}\:+\mathrm{1}}\:\:{rest}\:{decomposition}\:{of}\: \\ $$$${F}\left({t}\right)\:=\frac{{t}}{{t}^{\mathrm{4}} −\mathrm{2}{t}^{\mathrm{3}} \:+\mathrm{2}{t}^{\mathrm{2}} \:+\mathrm{6}{t}\:+\mathrm{1}}\:{the}\:{roots}\:{of}\:{p}\left({t}\right)={t}^{\mathrm{4}} −\mathrm{2}{t}^{\mathrm{3}} \:+\mathrm{2}{t}^{\mathrm{2}} \:+\mathrm{6}{t}\:+\mathrm{1} \\ $$$${are}\:{t}_{\mathrm{1}} =\mathrm{1},\mathrm{5898}\:+\mathrm{1},\mathrm{7445}{i}\:\left({complex}\right) \\ $$$${t}_{\mathrm{2}} =\mathrm{1},\mathrm{5898}−\mathrm{1},\mathrm{7445}{i}\left({complex}\right) \\ $$$${t}_{\mathrm{3}} =−\mathrm{1}\:\left({real}\right)\: \\ $$$${t}_{\mathrm{4}} =−\mathrm{0},\mathrm{1795}\left({real}\right)\:\Rightarrow{F}\left({t}\right)=\frac{{t}}{\left({t}+\mathrm{1}\right)\left({t}−{t}_{\mathrm{4}} \right)\left({t}−{t}_{\mathrm{1}} \right)\left({t}−\overset{−} {{t}}_{\mathrm{1}} \right)} \\ $$$$=\frac{{t}}{\left({t}+\mathrm{1}\right)\left({t}−{t}_{\mathrm{4}} \right)\left({t}^{\mathrm{2}} −\mathrm{2}{Re}\left({t}_{\mathrm{1}} \right){t}\:+\mid{t}_{\mathrm{1}} \mid^{\mathrm{2}} \right)}\:=\frac{{a}}{{t}+\mathrm{1}}\:+\frac{{b}}{{t}−{t}_{\mathrm{4}} }\:+\frac{{ct}\:+{d}}{{t}^{\mathrm{2}} −\mathrm{2}{Re}\left({t}_{\mathrm{1}} \right){t}\:+\mid{t}_{\mathrm{1}} \mid^{\mathrm{2}} } \\ $$$$...{be}\:{continued}... \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com