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Question Number 42370 by Raj Singh last updated on 24/Aug/18
Commented by maxmathsup by imad last updated on 24/Aug/18
let A = ∫ (dx/(sin(2x)+sinx)) A = ∫ (dx/(2sinx cosx +sinx)) = ∫ ((sinx)/(2sin^2 xcosx +sin^2 x)) dx = ∫ ((sinx)/(2(1−cos^2 x)cosx +1−cos^2 x)) = ∫ ((sinx)/(2cosx −2cos^3 x +1−cos^2 x))dx = ∫ ((sinx dx)/(−2cos^3 x −cos^2 x +2cosx +1)) =_(cosx =t) ∫ ((−dt)/(−2t^3 −t^2 +2t +1)) = ∫ (dt/(2t^3 +t^2 −2t−1)) but 2t^3 +t^2 −2t −1 = 2t^3 −2 +t^2 −2t +1 =2(t−1)(t^2 +t+1) +(t−1)^2 =(t−1)(2t^2 +2t +2 +t−1) =(t−1)(2t^2 +3t +1) roots of 2t^2 +3t +1 Δ =9−8 =1 ⇒t_1 =((−3+1)/4) =−(1/2) and t_2 =((−3−1)/4) =−1 ⇒ 2t^3 +t^2 −2t−1 =(t−1)(t+1)(t+(1/2)) let decompose F(t) =(1/(2t^3 +t^2 −2t −1)) =(1/((t−1)(t+1)(t+(1/2)))) F(t) =(a/(t−1)) +(b/(t+1)) +(c/(t+(1/2))) a =lim_(t→1) (t−1)F(t) = (1/(2.(3/2))) =(1/3) b =lim_(t→−1) (t+1)F(t) = (1/((−2)(−(1/2)))) =1 c =lim_(t→−(1/2)) (t+(1/2))F(t) = (1/((−(3/2))(1/2))) =−(4/3) ⇒ F(t) = (1/(3(t−1))) +(1/(t+1)) −(4/(3(t+(1/2)))) ⇒ A = ∫ F(t)dt =(1/3)ln∣t−1∣ +ln∣t+1∣ −(4/3)ln∣t +(1/2)∣ +c A =(1/3)ln∣cosx−1∣ +ln∣cosx +1∣ −(4/3)ln∣cosx +(1/2)∣ +c
$${let}\:{A}\:=\:\int\:\:\:\frac{{dx}}{{sin}\left(\mathrm{2}{x}\right)+{sinx}} \\ $$$${A}\:=\:\int\:\:\:\:\:\frac{{dx}}{\mathrm{2}{sinx}\:{cosx}\:+{sinx}}\:=\:\int\:\:\:\:\:\:\frac{{sinx}}{\mathrm{2}{sin}^{\mathrm{2}} {xcosx}\:+{sin}^{\mathrm{2}} {x}}\:{dx} \\ $$$$=\:\int\:\:\:\:\:\:\:\:\frac{{sinx}}{\mathrm{2}\left(\mathrm{1}−{cos}^{\mathrm{2}} {x}\right){cosx}\:+\mathrm{1}−{cos}^{\mathrm{2}} {x}}\:=\:\int\:\:\:\:\frac{{sinx}}{\mathrm{2}{cosx}\:−\mathrm{2}{cos}^{\mathrm{3}} {x}\:+\mathrm{1}−{cos}^{\mathrm{2}} {x}}{dx} \\ $$$$=\:\int\:\:\:\:\:\:\:\:\:\frac{{sinx}\:{dx}}{−\mathrm{2}{cos}^{\mathrm{3}} {x}\:−{cos}^{\mathrm{2}} {x}\:+\mathrm{2}{cosx}\:+\mathrm{1}}\:=_{{cosx}\:={t}} \:\:\:\:\int\:\:\:\:\:\:\:\:\frac{−{dt}}{−\mathrm{2}{t}^{\mathrm{3}} −{t}^{\mathrm{2}} \:+\mathrm{2}{t}\:+\mathrm{1}} \\ $$$$=\:\int\:\:\:\:\:\:\frac{{dt}}{\mathrm{2}{t}^{\mathrm{3}} \:+{t}^{\mathrm{2}} \:−\mathrm{2}{t}−\mathrm{1}}\:\:{but}\:\:\mathrm{2}{t}^{\mathrm{3}} \:+{t}^{\mathrm{2}} −\mathrm{2}{t}\:−\mathrm{1}\:=\:\mathrm{2}{t}^{\mathrm{3}} −\mathrm{2}\:+{t}^{\mathrm{2}} \:−\mathrm{2}{t}\:+\mathrm{1} \\ $$$$=\mathrm{2}\left({t}−\mathrm{1}\right)\left({t}^{\mathrm{2}} \:+{t}+\mathrm{1}\right)\:+\left({t}−\mathrm{1}\right)^{\mathrm{2}} \:=\left({t}−\mathrm{1}\right)\left(\mathrm{2}{t}^{\mathrm{2}} \:+\mathrm{2}{t}\:+\mathrm{2}\:+{t}−\mathrm{1}\right) \\ $$$$=\left({t}−\mathrm{1}\right)\left(\mathrm{2}{t}^{\mathrm{2}} \:\:+\mathrm{3}{t}\:+\mathrm{1}\right)\:\:{roots}\:{of}\:\:\mathrm{2}{t}^{\mathrm{2}} \:+\mathrm{3}{t}\:+\mathrm{1}\: \\ $$$$\Delta\:=\mathrm{9}−\mathrm{8}\:=\mathrm{1}\:\Rightarrow{t}_{\mathrm{1}} =\frac{−\mathrm{3}+\mathrm{1}}{\mathrm{4}}\:=−\frac{\mathrm{1}}{\mathrm{2}}\:\:{and}\:{t}_{\mathrm{2}} =\frac{−\mathrm{3}−\mathrm{1}}{\mathrm{4}}\:=−\mathrm{1}\:\Rightarrow \\ $$$$\mathrm{2}{t}^{\mathrm{3}} \:+{t}^{\mathrm{2}} \:−\mathrm{2}{t}−\mathrm{1}\:=\left({t}−\mathrm{1}\right)\left({t}+\mathrm{1}\right)\left({t}+\frac{\mathrm{1}}{\mathrm{2}}\right)\:{let}\:{decompose} \\ $$$${F}\left({t}\right)\:=\frac{\mathrm{1}}{\mathrm{2}{t}^{\mathrm{3}} \:+{t}^{\mathrm{2}} −\mathrm{2}{t}\:−\mathrm{1}}\:=\frac{\mathrm{1}}{\left({t}−\mathrm{1}\right)\left({t}+\mathrm{1}\right)\left({t}+\frac{\mathrm{1}}{\mathrm{2}}\right)} \\ $$$${F}\left({t}\right)\:=\frac{{a}}{{t}−\mathrm{1}}\:+\frac{{b}}{{t}+\mathrm{1}}\:+\frac{{c}}{{t}+\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$${a}\:={lim}_{{t}\rightarrow\mathrm{1}} \left({t}−\mathrm{1}\right){F}\left({t}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}.\frac{\mathrm{3}}{\mathrm{2}}}\:=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${b}\:={lim}_{{t}\rightarrow−\mathrm{1}} \left({t}+\mathrm{1}\right){F}\left({t}\right)\:=\:\frac{\mathrm{1}}{\left(−\mathrm{2}\right)\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)}\:=\mathrm{1} \\ $$$${c}\:={lim}_{{t}\rightarrow−\frac{\mathrm{1}}{\mathrm{2}}} \left({t}+\frac{\mathrm{1}}{\mathrm{2}}\right){F}\left({t}\right)\:=\:\frac{\mathrm{1}}{\left(−\frac{\mathrm{3}}{\mathrm{2}}\right)\frac{\mathrm{1}}{\mathrm{2}}}\:=−\frac{\mathrm{4}}{\mathrm{3}}\:\Rightarrow \\ $$$${F}\left({t}\right)\:=\:\frac{\mathrm{1}}{\mathrm{3}\left({t}−\mathrm{1}\right)}\:+\frac{\mathrm{1}}{{t}+\mathrm{1}}\:\:\:−\frac{\mathrm{4}}{\mathrm{3}\left({t}+\frac{\mathrm{1}}{\mathrm{2}}\right)}\:\Rightarrow \\ $$$${A}\:\:=\:\int\:\:{F}\left({t}\right){dt}\:\:=\frac{\mathrm{1}}{\mathrm{3}}{ln}\mid{t}−\mathrm{1}\mid\:+{ln}\mid{t}+\mathrm{1}\mid\:−\frac{\mathrm{4}}{\mathrm{3}}{ln}\mid{t}\:+\frac{\mathrm{1}}{\mathrm{2}}\mid\:+{c} \\ $$$${A}\:=\frac{\mathrm{1}}{\mathrm{3}}{ln}\mid{cosx}−\mathrm{1}\mid\:+{ln}\mid{cosx}\:+\mathrm{1}\mid\:−\frac{\mathrm{4}}{\mathrm{3}}{ln}\mid{cosx}\:+\frac{\mathrm{1}}{\mathrm{2}}\mid\:+{c} \\ $$$$ \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 24/Aug/18
∫(dx/(sinx(2cosx+1))) ∫((sinxdx)/((1−cos^2 x)(2cosx+1))) t=cosx dt=−sinxdx ∫((−dt)/((1−t^2 )(2t+1))) ∫(dt/((t+1)(t−1)(2t+1))) (1/((t+1)(t−1)(2t+1)))=(a/(t+1))+(b/(t−1))+(c/(2t+1)) 1=a(t−1)(2t+1)+b(t+1)(2t+1)+c(t+1)(t−1) put t−1=0 t=1 b(1+1)(2+1)=1 b=(1/6) put t+1=0 a(−1−1)(2×−1+1)=1 a(−2)(−1)=1 a=(1/2) put 2t+1=0 c(((−1)/2)+1)(((−1)/2)−1)=1 c((1/4)−1)=1 c×((−3)/4)=1 c=((−4)/3) ∫(a/(t+1))dt+∫(b/(t−1))dt+∫(c/(2t+1))dt a∫(dt/(t+1))+b∫(dt/(t−1))+(c/2)∫(dt/(t+(1/2))) aln(t+1)+bln(t−1)+(c/2)ln(t+(1/2)) (1/2)ln(cosx+1)+(1/6)ln(cosx−1)+((−4)/(2×3))ln(cosx+(1/2))+k
$$\int\frac{{dx}}{{sinx}\left(\mathrm{2}{cosx}+\mathrm{1}\right)} \\ $$$$\int\frac{{sinxdx}}{\left(\mathrm{1}−{cos}^{\mathrm{2}} {x}\right)\left(\mathrm{2}{cosx}+\mathrm{1}\right)} \\ $$$${t}={cosx}\:\:\:{dt}=−{sinxdx} \\ $$$$\int\frac{−{dt}}{\left(\mathrm{1}−{t}^{\mathrm{2}} \right)\left(\mathrm{2}{t}+\mathrm{1}\right)} \\ $$$$\int\frac{{dt}}{\left({t}+\mathrm{1}\right)\left({t}−\mathrm{1}\right)\left(\mathrm{2}{t}+\mathrm{1}\right)} \\ $$$$\frac{\mathrm{1}}{\left({t}+\mathrm{1}\right)\left({t}−\mathrm{1}\right)\left(\mathrm{2}{t}+\mathrm{1}\right)}=\frac{{a}}{{t}+\mathrm{1}}+\frac{{b}}{{t}−\mathrm{1}}+\frac{{c}}{\mathrm{2}{t}+\mathrm{1}} \\ $$$$\mathrm{1}={a}\left({t}−\mathrm{1}\right)\left(\mathrm{2}{t}+\mathrm{1}\right)+{b}\left({t}+\mathrm{1}\right)\left(\mathrm{2}{t}+\mathrm{1}\right)+{c}\left({t}+\mathrm{1}\right)\left({t}−\mathrm{1}\right) \\ $$$${put}\:{t}−\mathrm{1}=\mathrm{0}\:\:\:\:{t}=\mathrm{1} \\ $$$${b}\left(\mathrm{1}+\mathrm{1}\right)\left(\mathrm{2}+\mathrm{1}\right)=\mathrm{1}\:\:\:\:\boldsymbol{{b}}=\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$\boldsymbol{{put}}\:\boldsymbol{{t}}+\mathrm{1}=\mathrm{0} \\ $$$$\boldsymbol{{a}}\left(−\mathrm{1}−\mathrm{1}\right)\left(\mathrm{2}×−\mathrm{1}+\mathrm{1}\right)=\mathrm{1} \\ $$$$\boldsymbol{{a}}\left(−\mathrm{2}\right)\left(−\mathrm{1}\right)=\mathrm{1}\:\:\:\:\boldsymbol{{a}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\boldsymbol{{put}}\:\mathrm{2}\boldsymbol{{t}}+\mathrm{1}=\mathrm{0} \\ $$$$\boldsymbol{{c}}\left(\frac{−\mathrm{1}}{\mathrm{2}}+\mathrm{1}\right)\left(\frac{−\mathrm{1}}{\mathrm{2}}−\mathrm{1}\right)=\mathrm{1} \\ $$$$\boldsymbol{{c}}\left(\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}\right)=\mathrm{1} \\ $$$$\boldsymbol{{c}}×\frac{−\mathrm{3}}{\mathrm{4}}=\mathrm{1}\:\:\:\:\:\boldsymbol{{c}}=\frac{−\mathrm{4}}{\mathrm{3}} \\ $$$$\int\frac{{a}}{{t}+\mathrm{1}}{dt}+\int\frac{{b}}{{t}−\mathrm{1}}{dt}+\int\frac{{c}}{\mathrm{2}{t}+\mathrm{1}}{dt} \\ $$$${a}\int\frac{{dt}}{{t}+\mathrm{1}}+{b}\int\frac{{dt}}{{t}−\mathrm{1}}+\frac{{c}}{\mathrm{2}}\int\frac{{dt}}{{t}+\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$${aln}\left({t}+\mathrm{1}\right)+{bln}\left({t}−\mathrm{1}\right)+\frac{{c}}{\mathrm{2}}{ln}\left({t}+\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({cosx}+\mathrm{1}\right)+\frac{\mathrm{1}}{\mathrm{6}}{ln}\left({cosx}−\mathrm{1}\right)+\frac{−\mathrm{4}}{\mathrm{2}×\mathrm{3}}{ln}\left({cosx}+\frac{\mathrm{1}}{\mathrm{2}}\right)+{k} \\ $$$$ \\ $$$$ \\ $$

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