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Question Number 62266 by aliesam last updated on 18/Jun/19
∫((2sin(x)+3cos(x))/(3sin(x)+4cos(x)))dx
$$\int\frac{\mathrm{2}{sin}\left({x}\right)+\mathrm{3}{cos}\left({x}\right)}{\mathrm{3}{sin}\left({x}\right)+\mathrm{4}{cos}\left({x}\right)}{dx} \\ $$
Commented by maxmathsup by imad last updated on 19/Jun/19
let A =∫ ((2sinx +3cosx)/(3sinx +4cosx)) dx changement tan((x/2))=t give A =∫ ((((4t)/(1+t^2 ))+3((1−t^2 )/(1+t^2 )))/(((6t)/(1+t^2 )) +4((1−t^2 )/(1+t^2 )))) ((2dt)/(1+t^2 )) = ∫ ((4t+3−3t^2 )/((6t+4−4t^2 )(1+t^2 ))) ((2dt)/(1+t^2 )) = ∫ ((−3t^2 +4t+3)/((1+t^2 )(−2t^2 +3t+2))) dt = ∫ ((3t^2 −4t−3)/((t^2 +1)(2t^2 −3t−2)))dt let decompose F(t) =((3t^2 −4t−3)/((t^2 +1)(2t^2 −3t−2))) 2t^2 −3t −2 =0 →Δ =9−4(−4) =9+16 =25 ⇒t_1 =((3+5)/4) =2 , t_2 =((3−5)/4) =−(1/2) ⇒ F(t) =((3t^2 −4t−3)/((t−2)(t+(1/2))(t^2 +1))) = ((2(3t^2 −4t−3))/((t−2)(2t+1)(t^2 +1))) =(a/(t−2)) +(b/(2t+1)) +((ct+d)/(t^2 +1)) a =lim_(t→2) (t−2)F(t) =((2(12−8−3))/(5.5)) =(2/(25)) b =lim_(t→−(1/2)) (2t+1)F(t) =((2((3/4) +2−3))/(−(5/2).(5/4))) =((2(3+8−12))/(4(−(5/2)).(5/4))) =(4/(25)) ⇒ F(t) = (2/(25(t−2))) +(4/(25(2t+1))) +((ct +d)/(t^2 +1)) lim_(t→+∞) tF(t) =0 = a+(b/2) +c ⇒c =−a−(b/2) =−(2/(25)) −(2/(25)) =−(4/(25)) ⇒ F(t) =(2/(25(t−2))) +(4/(25(2t+1))) +((−(4/(25))t +d)/(t^2 +1)) F(0) =(3/2) =−(1/(25)) +(4/(25)) +d =(3/(25)) +d ⇒d =(3/2)−(3/(25)) =3((1/2) −(1/(25)))=3.((23)/(50)) =((69)/(50)) ⇒ ∫ F(t)dt = (2/(25)) ∫(dt/(t−2)) +(4/(25)) ∫ (dt/(2t+1)) −(1/(50)) ∫ ((8t−69)/(t^2 +1))dt +c =(2/(25))ln∣t−2∣+(4/(50))ln∣2t+1∣ −(4/(50)) ∫ ((2t)/(t^2 +1)) dt +((69)/(50)) ∫ (dt/(t^2 +1)) dt +c =(2/(25))ln∣t−2∣+(4/(50))ln∣2t+1∣−(4/(50))ln(t^2 +1) +((69)/(50)) arctan(t)+c ⇒ A =(2/(25))ln∣tan((x/2))−2∣+(4/(50))ln∣2tan((x/2))+1∣−(4/(50))ln(1+tan^2 ((x/2))) +((69)/(100)) x +c .
$${let}\:{A}\:=\int\:\:\:\frac{\mathrm{2}{sinx}\:+\mathrm{3}{cosx}}{\mathrm{3}{sinx}\:+\mathrm{4}{cosx}}\:{dx}\:{changement}\:{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}\:{give} \\ $$$${A}\:=\int\:\:\:\frac{\frac{\mathrm{4}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }+\mathrm{3}\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}{\frac{\mathrm{6}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }\:+\mathrm{4}\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }\:=\:\int\:\:\frac{\mathrm{4}{t}+\mathrm{3}−\mathrm{3}{t}^{\mathrm{2}} }{\left(\mathrm{6}{t}+\mathrm{4}−\mathrm{4}{t}^{\mathrm{2}} \right)\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$=\:\int\:\:\:\frac{−\mathrm{3}{t}^{\mathrm{2}} \:+\mathrm{4}{t}+\mathrm{3}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left(−\mathrm{2}{t}^{\mathrm{2}} \:+\mathrm{3}{t}+\mathrm{2}\right)}\:{dt}\:=\:\int\:\:\frac{\mathrm{3}{t}^{\mathrm{2}} −\mathrm{4}{t}−\mathrm{3}}{\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)\left(\mathrm{2}{t}^{\mathrm{2}} −\mathrm{3}{t}−\mathrm{2}\right)}{dt}\:\:{let}\:{decompose} \\ $$$${F}\left({t}\right)\:=\frac{\mathrm{3}{t}^{\mathrm{2}} −\mathrm{4}{t}−\mathrm{3}}{\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)\left(\mathrm{2}{t}^{\mathrm{2}} −\mathrm{3}{t}−\mathrm{2}\right)} \\ $$$$\mathrm{2}{t}^{\mathrm{2}} −\mathrm{3}{t}\:−\mathrm{2}\:=\mathrm{0}\:\rightarrow\Delta\:=\mathrm{9}−\mathrm{4}\left(−\mathrm{4}\right)\:=\mathrm{9}+\mathrm{16}\:=\mathrm{25}\:\Rightarrow{t}_{\mathrm{1}} =\frac{\mathrm{3}+\mathrm{5}}{\mathrm{4}}\:=\mathrm{2}\:\:,\:{t}_{\mathrm{2}} =\frac{\mathrm{3}−\mathrm{5}}{\mathrm{4}}\:=−\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow \\ $$$${F}\left({t}\right)\:=\frac{\mathrm{3}{t}^{\mathrm{2}} −\mathrm{4}{t}−\mathrm{3}}{\left({t}−\mathrm{2}\right)\left({t}+\frac{\mathrm{1}}{\mathrm{2}}\right)\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)}\:=\:\frac{\mathrm{2}\left(\mathrm{3}{t}^{\mathrm{2}} −\mathrm{4}{t}−\mathrm{3}\right)}{\left({t}−\mathrm{2}\right)\left(\mathrm{2}{t}+\mathrm{1}\right)\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)}\:=\frac{{a}}{{t}−\mathrm{2}}\:+\frac{{b}}{\mathrm{2}{t}+\mathrm{1}}\:+\frac{{ct}+{d}}{{t}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${a}\:={lim}_{{t}\rightarrow\mathrm{2}} \:\:\:\left({t}−\mathrm{2}\right){F}\left({t}\right)\:=\frac{\mathrm{2}\left(\mathrm{12}−\mathrm{8}−\mathrm{3}\right)}{\mathrm{5}.\mathrm{5}}\:=\frac{\mathrm{2}}{\mathrm{25}} \\ $$$${b}\:={lim}_{{t}\rightarrow−\frac{\mathrm{1}}{\mathrm{2}}} \:\:\:\left(\mathrm{2}{t}+\mathrm{1}\right){F}\left({t}\right)\:=\frac{\mathrm{2}\left(\frac{\mathrm{3}}{\mathrm{4}}\:+\mathrm{2}−\mathrm{3}\right)}{−\frac{\mathrm{5}}{\mathrm{2}}.\frac{\mathrm{5}}{\mathrm{4}}}\:=\frac{\mathrm{2}\left(\mathrm{3}+\mathrm{8}−\mathrm{12}\right)}{\mathrm{4}\left(−\frac{\mathrm{5}}{\mathrm{2}}\right).\frac{\mathrm{5}}{\mathrm{4}}}\:=\frac{\mathrm{4}}{\mathrm{25}}\:\Rightarrow \\ $$$${F}\left({t}\right)\:=\:\frac{\mathrm{2}}{\mathrm{25}\left({t}−\mathrm{2}\right)}\:+\frac{\mathrm{4}}{\mathrm{25}\left(\mathrm{2}{t}+\mathrm{1}\right)}\:+\frac{{ct}\:+{d}}{{t}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${lim}_{{t}\rightarrow+\infty} \:{tF}\left({t}\right)\:=\mathrm{0}\:=\:{a}+\frac{{b}}{\mathrm{2}}\:+{c}\:\:\Rightarrow{c}\:=−{a}−\frac{{b}}{\mathrm{2}}\:=−\frac{\mathrm{2}}{\mathrm{25}}\:−\frac{\mathrm{2}}{\mathrm{25}}\:=−\frac{\mathrm{4}}{\mathrm{25}}\:\Rightarrow \\ $$$${F}\left({t}\right)\:=\frac{\mathrm{2}}{\mathrm{25}\left({t}−\mathrm{2}\right)}\:+\frac{\mathrm{4}}{\mathrm{25}\left(\mathrm{2}{t}+\mathrm{1}\right)}\:+\frac{−\frac{\mathrm{4}}{\mathrm{25}}{t}\:+{d}}{{t}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${F}\left(\mathrm{0}\right)\:=\frac{\mathrm{3}}{\mathrm{2}}\:=−\frac{\mathrm{1}}{\mathrm{25}}\:+\frac{\mathrm{4}}{\mathrm{25}}\:+{d}\:=\frac{\mathrm{3}}{\mathrm{25}}\:+{d}\:\Rightarrow{d}\:=\frac{\mathrm{3}}{\mathrm{2}}−\frac{\mathrm{3}}{\mathrm{25}}\:=\mathrm{3}\left(\frac{\mathrm{1}}{\mathrm{2}}\:−\frac{\mathrm{1}}{\mathrm{25}}\right)=\mathrm{3}.\frac{\mathrm{23}}{\mathrm{50}}\:=\frac{\mathrm{69}}{\mathrm{50}}\:\Rightarrow \\ $$$$\int\:{F}\left({t}\right){dt}\:=\:\frac{\mathrm{2}}{\mathrm{25}}\:\int\frac{{dt}}{{t}−\mathrm{2}}\:+\frac{\mathrm{4}}{\mathrm{25}}\:\int\:\frac{{dt}}{\mathrm{2}{t}+\mathrm{1}}\:−\frac{\mathrm{1}}{\mathrm{50}}\:\int\:\frac{\mathrm{8}{t}−\mathrm{69}}{{t}^{\mathrm{2}} \:+\mathrm{1}}{dt}\:+{c} \\ $$$$=\frac{\mathrm{2}}{\mathrm{25}}{ln}\mid{t}−\mathrm{2}\mid+\frac{\mathrm{4}}{\mathrm{50}}{ln}\mid\mathrm{2}{t}+\mathrm{1}\mid\:−\frac{\mathrm{4}}{\mathrm{50}}\:\int\:\:\frac{\mathrm{2}{t}}{{t}^{\mathrm{2}} \:+\mathrm{1}}\:{dt}\:+\frac{\mathrm{69}}{\mathrm{50}}\:\int\:\frac{{dt}}{{t}^{\mathrm{2}} \:+\mathrm{1}}\:{dt}\:+{c} \\ $$$$=\frac{\mathrm{2}}{\mathrm{25}}{ln}\mid{t}−\mathrm{2}\mid+\frac{\mathrm{4}}{\mathrm{50}}{ln}\mid\mathrm{2}{t}+\mathrm{1}\mid−\frac{\mathrm{4}}{\mathrm{50}}{ln}\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)\:+\frac{\mathrm{69}}{\mathrm{50}}\:{arctan}\left({t}\right)+{c}\:\Rightarrow \\ $$$${A}\:=\frac{\mathrm{2}}{\mathrm{25}}{ln}\mid{tan}\left(\frac{{x}}{\mathrm{2}}\right)−\mathrm{2}\mid+\frac{\mathrm{4}}{\mathrm{50}}{ln}\mid\mathrm{2}{tan}\left(\frac{{x}}{\mathrm{2}}\right)+\mathrm{1}\mid−\frac{\mathrm{4}}{\mathrm{50}}{ln}\left(\mathrm{1}+{tan}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)\right)\:+\frac{\mathrm{69}}{\mathrm{100}}\:{x}\:+{c}\:. \\ $$$$ \\ $$$$ \\ $$
Answered by tanmay last updated on 18/Jun/19
N_r =a(D_r )+b((dD_r /dx)) ∫((a(D_r )+b((dD_r /dx)))/D_r )dx a∫dx+b∫((d(D_r ))/D_r ) ax+bln(D_r )+c ax+bln(3sinx+4cosx)+c now finding value of a and b 2sinx+3cosx=a(3sinx+4cosx)+b(d/dx)(3sinx+4cosx) 2sinx+3cosx=a(3sinx+4cosx)+b(3cosx−4sinx) 2sinx+3cosx=(3a−4b)sinx+(4a+3b)cosx 3a−4b=2 ×3 4a+3b=3 ×4 9a−12b=6 16a+12b=12 25a=18→a=((18)/(25)) 3b=3−4a 3b=3−((72)/(25))→b=(1/(25)) so answer is ((18)/(25))x+(1/(25))ln(3sinx+4cosx)+c
$${N}_{{r}} ={a}\left({D}_{{r}} \right)+{b}\left(\frac{{dD}_{{r}} }{{dx}}\right) \\ $$$$\int\frac{{a}\left({D}_{{r}} \right)+{b}\left(\frac{{dD}_{{r}} }{{dx}}\right)}{{D}_{{r}} }{dx} \\ $$$${a}\int{dx}+{b}\int\frac{{d}\left({D}_{{r}} \right)}{{D}_{{r}} } \\ $$$${ax}+{bln}\left({D}_{{r}} \right)+{c} \\ $$$${ax}+{bln}\left(\mathrm{3}{sinx}+\mathrm{4}{cosx}\right)+{c} \\ $$$${now}\:{finding}\:{value}\:{of}\:{a}\:{and}\:{b} \\ $$$$\mathrm{2}{sinx}+\mathrm{3}{cosx}={a}\left(\mathrm{3}{sinx}+\mathrm{4}{cosx}\right)+{b}\frac{{d}}{{dx}}\left(\mathrm{3}{sinx}+\mathrm{4}{cosx}\right) \\ $$$$\mathrm{2}{sinx}+\mathrm{3}{cosx}={a}\left(\mathrm{3}{sinx}+\mathrm{4}{cosx}\right)+{b}\left(\mathrm{3}{cosx}−\mathrm{4}{sinx}\right) \\ $$$$\mathrm{2}{sinx}+\mathrm{3}{cosx}=\left(\mathrm{3}{a}−\mathrm{4}{b}\right){sinx}+\left(\mathrm{4}{a}+\mathrm{3}{b}\right){cosx} \\ $$$$\mathrm{3}{a}−\mathrm{4}{b}=\mathrm{2}\:\:×\mathrm{3} \\ $$$$\mathrm{4}{a}+\mathrm{3}{b}=\mathrm{3}\:\:\:×\mathrm{4} \\ $$$$\mathrm{9}{a}−\mathrm{12}{b}=\mathrm{6} \\ $$$$\mathrm{16}{a}+\mathrm{12}{b}=\mathrm{12} \\ $$$$\mathrm{25}{a}=\mathrm{18}\rightarrow{a}=\frac{\mathrm{18}}{\mathrm{25}} \\ $$$$\mathrm{3}{b}=\mathrm{3}−\mathrm{4}{a} \\ $$$$\mathrm{3}{b}=\mathrm{3}−\frac{\mathrm{72}}{\mathrm{25}}\rightarrow{b}=\frac{\mathrm{1}}{\mathrm{25}} \\ $$$$\boldsymbol{{so}}\:\boldsymbol{{answer}}\:\boldsymbol{{is}}\: \\ $$$$\frac{\mathrm{18}}{\mathrm{25}}\boldsymbol{{x}}+\frac{\mathrm{1}}{\mathrm{25}}\boldsymbol{{ln}}\left(\mathrm{3}\boldsymbol{{sinx}}+\mathrm{4}\boldsymbol{{cosx}}\right)+\boldsymbol{{c}} \\ $$
Commented by aliesam last updated on 18/Jun/19
thank you sir
$${thank}\:{you}\:{sir}\: \\ $$
Commented by tanmay last updated on 18/Jun/19
most welcome sir
$${most}\:{welcome}\:{sir} \\ $$

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