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Question Number 53383 by gunawan last updated on 21/Jan/19

If ∫(1/((sin x+4)(sin x−1)))dx = A(1/(tan (x/2)−1))+B tan^(−1) (f(x))+C, then

$$\mathrm{If}\:\int\frac{\mathrm{1}}{\left(\mathrm{sin}\:{x}+\mathrm{4}\right)\left(\mathrm{sin}\:{x}−\mathrm{1}\right)}{dx} \\ $$$$\:\:\:\:\:\:\:\:\:=\:{A}\frac{\mathrm{1}}{\mathrm{tan}\:\frac{{x}}{\mathrm{2}}−\mathrm{1}}+{B}\:\mathrm{tan}^{−\mathrm{1}} \left({f}\left({x}\right)\right)+{C},\:\mathrm{then} \\ $$

Commented by maxmathsup by imad last updated on 21/Jan/19

let A =∫ (dx/((sinx+4)(sinx−1))) ⇒A =_(tan((x/2))=t) ∫ (1/((((2t)/(1+t^2 )) +4)(((2t)/(1+t^2 ))−1))) ((2dt)/(1+t^2 )) = ∫ (2/((4t^2 +4+2t)(2t−1−t^2 ))) (((1+t^2 )^2 )/(1+t^2 ))dt =2∫ ((1+t^2 )/((4t^2 +2t+4)(−t^2 +2t−1))) dt =−∫ ((1+t^2 )/((2t^2 +t+2)(t−1)^2 ))dt let decompose F(t)=((t^2 +1)/((2t^2 +t+2)(t−1)^2 )) ⇒F(t)=(a/(t−1)) +(b/((t−1)^2 )) +((ct +d)/(2t^2 +t+2)) b=lim_(t→1) (t−1)^2 F(t)=(2/5) lim_(t→+∞) tF(t)=0 =a +(c/2) ⇒c=−2a ⇒ F(t)=(a/(t−1)) +(2/(5(t−1)^2 )) +((−2at +d)/(2t^2 +t+2)) F(0)=(1/2) =−a +(2/5) +(d/2) ⇒1 =−2a+(4/5) +d ⇒−2a+d=1−(4/5) =(1/5) F(2)=(5/(12)) =a +(2/5) +((−4a+d)/(12)) ⇒5=12a +((24)/5) −4a+d ⇒5−((24)/5) =8a +d ⇒ 8a+d =(1/5) ⇒d=(1/5) −8a ⇒−2a+(1/5) −8a =(1/5) ⇒a=0 ⇒d=(1/5) ⇒ F(t)=(2/(5(t−1)^2 )) +(1/(5(2t^2 +t+2))) ⇒A =−(2/5) ∫ (dt/((t−1)^2 )) −(1/5) ∫ (dt/(2t^2 +t+2)) ∫ (dt/((t−1)^2 )) =−(1/(t−1)) +c_1 ∫ (dt/(2t^2 +t +2)) =(1/2) ∫ (dt/(t^2 +2(1/4)t +(1/(16))+1−(1/(16)))) =(1/2) ∫ (dt/((t+(1/4))^2 +((15)/(16)))) =_(t+(1/4)=((√(15))/4)u) (1/2) ∫ (1/(((15)/(16))(1+u^2 ))) ((√(15))/4) du=(1/2) .((16)/(15)) .((√(15))/4) arctan(u)+c_2 =(2/(√(15))) arctan(((4t+1)/(√(15)))) +c_2 ⇒A=(2/5) (1/(tan((x/2))−1)) −(1/(5(√(15)))) arctan(((4tan((x/2))+1)/(√(15)))) +C ⇒A =(2/5) , B =−(1/(5(√(15)))) , f(x)=((4arctan((x/2))+1)/(√(15))) .

$${let}\:{A}\:=\int\:\:\:\:\frac{{dx}}{\left({sinx}+\mathrm{4}\right)\left({sinx}−\mathrm{1}\right)}\:\Rightarrow{A}\:=_{{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}} \:\:\:\int\:\:\:\frac{\mathrm{1}}{\left(\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }\:+\mathrm{4}\right)\left(\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }−\mathrm{1}\right)}\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$=\:\int\:\:\:\:\frac{\mathrm{2}}{\left(\mathrm{4}{t}^{\mathrm{2}} +\mathrm{4}+\mathrm{2}{t}\right)\left(\mathrm{2}{t}−\mathrm{1}−{t}^{\mathrm{2}} \right)}\:\frac{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$$=\mathrm{2}\int\:\:\:\:\:\frac{\mathrm{1}+{t}^{\mathrm{2}} }{\left(\mathrm{4}{t}^{\mathrm{2}} \:+\mathrm{2}{t}+\mathrm{4}\right)\left(−{t}^{\mathrm{2}} +\mathrm{2}{t}−\mathrm{1}\right)}\:{dt}\:=−\int\:\:\:\frac{\mathrm{1}+{t}^{\mathrm{2}} }{\left(\mathrm{2}{t}^{\mathrm{2}} \:+{t}+\mathrm{2}\right)\left({t}−\mathrm{1}\right)^{\mathrm{2}} }{dt} \\ $$$${let}\:{decompose}\:{F}\left({t}\right)=\frac{{t}^{\mathrm{2}} \:+\mathrm{1}}{\left(\mathrm{2}{t}^{\mathrm{2}} \:+{t}+\mathrm{2}\right)\left({t}−\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow{F}\left({t}\right)=\frac{{a}}{{t}−\mathrm{1}}\:+\frac{{b}}{\left({t}−\mathrm{1}\right)^{\mathrm{2}} }\:+\frac{{ct}\:+{d}}{\mathrm{2}{t}^{\mathrm{2}} \:+{t}+\mathrm{2}} \\ $$$${b}={lim}_{{t}\rightarrow\mathrm{1}} \left({t}−\mathrm{1}\right)^{\mathrm{2}} {F}\left({t}\right)=\frac{\mathrm{2}}{\mathrm{5}} \\ $$$${lim}_{{t}\rightarrow+\infty} {tF}\left({t}\right)=\mathrm{0}\:={a}\:+\frac{{c}}{\mathrm{2}}\:\Rightarrow{c}=−\mathrm{2}{a}\:\Rightarrow \\ $$$${F}\left({t}\right)=\frac{{a}}{{t}−\mathrm{1}}\:+\frac{\mathrm{2}}{\mathrm{5}\left({t}−\mathrm{1}\right)^{\mathrm{2}} }\:+\frac{−\mathrm{2}{at}\:+{d}}{\mathrm{2}{t}^{\mathrm{2}} \:+{t}+\mathrm{2}} \\ $$$${F}\left(\mathrm{0}\right)=\frac{\mathrm{1}}{\mathrm{2}}\:=−{a}\:+\frac{\mathrm{2}}{\mathrm{5}}\:+\frac{{d}}{\mathrm{2}}\:\Rightarrow\mathrm{1}\:=−\mathrm{2}{a}+\frac{\mathrm{4}}{\mathrm{5}}\:+{d}\:\Rightarrow−\mathrm{2}{a}+{d}=\mathrm{1}−\frac{\mathrm{4}}{\mathrm{5}}\:=\frac{\mathrm{1}}{\mathrm{5}} \\ $$$${F}\left(\mathrm{2}\right)=\frac{\mathrm{5}}{\mathrm{12}}\:={a}\:+\frac{\mathrm{2}}{\mathrm{5}}\:+\frac{−\mathrm{4}{a}+{d}}{\mathrm{12}}\:\Rightarrow\mathrm{5}=\mathrm{12}{a}\:+\frac{\mathrm{24}}{\mathrm{5}}\:−\mathrm{4}{a}+{d}\:\Rightarrow\mathrm{5}−\frac{\mathrm{24}}{\mathrm{5}}\:=\mathrm{8}{a}\:+{d}\:\Rightarrow \\ $$$$\mathrm{8}{a}+{d}\:=\frac{\mathrm{1}}{\mathrm{5}}\:\Rightarrow{d}=\frac{\mathrm{1}}{\mathrm{5}}\:−\mathrm{8}{a}\:\Rightarrow−\mathrm{2}{a}+\frac{\mathrm{1}}{\mathrm{5}}\:−\mathrm{8}{a}\:=\frac{\mathrm{1}}{\mathrm{5}}\:\Rightarrow{a}=\mathrm{0}\:\Rightarrow{d}=\frac{\mathrm{1}}{\mathrm{5}}\:\Rightarrow \\ $$$${F}\left({t}\right)=\frac{\mathrm{2}}{\mathrm{5}\left({t}−\mathrm{1}\right)^{\mathrm{2}} }\:+\frac{\mathrm{1}}{\mathrm{5}\left(\mathrm{2}{t}^{\mathrm{2}} \:+{t}+\mathrm{2}\right)}\:\Rightarrow{A}\:=−\frac{\mathrm{2}}{\mathrm{5}}\:\int\:\frac{{dt}}{\left({t}−\mathrm{1}\right)^{\mathrm{2}} }\:−\frac{\mathrm{1}}{\mathrm{5}}\:\int\:\:\frac{{dt}}{\mathrm{2}{t}^{\mathrm{2}} \:+{t}+\mathrm{2}} \\ $$$$\int\:\:\frac{{dt}}{\left({t}−\mathrm{1}\right)^{\mathrm{2}} }\:=−\frac{\mathrm{1}}{{t}−\mathrm{1}}\:+{c}_{\mathrm{1}} \\ $$$$\int\:\:\:\frac{{dt}}{\mathrm{2}{t}^{\mathrm{2}} \:+{t}\:+\mathrm{2}}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\int\:\:\frac{{dt}}{{t}^{\mathrm{2}} \:+\mathrm{2}\frac{\mathrm{1}}{\mathrm{4}}{t}\:\:+\frac{\mathrm{1}}{\mathrm{16}}+\mathrm{1}−\frac{\mathrm{1}}{\mathrm{16}}}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\int\:\:\frac{{dt}}{\left({t}+\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{2}} \:+\frac{\mathrm{15}}{\mathrm{16}}} \\ $$$$=_{{t}+\frac{\mathrm{1}}{\mathrm{4}}=\frac{\sqrt{\mathrm{15}}}{\mathrm{4}}{u}} \:\frac{\mathrm{1}}{\mathrm{2}}\:\:\int\:\:\:\frac{\mathrm{1}}{\frac{\mathrm{15}}{\mathrm{16}}\left(\mathrm{1}+{u}^{\mathrm{2}} \right)}\:\frac{\sqrt{\mathrm{15}}}{\mathrm{4}}\:{du}=\frac{\mathrm{1}}{\mathrm{2}}\:.\frac{\mathrm{16}}{\mathrm{15}}\:.\frac{\sqrt{\mathrm{15}}}{\mathrm{4}}\:{arctan}\left({u}\right)+{c}_{\mathrm{2}} \\ $$$$=\frac{\mathrm{2}}{\sqrt{\mathrm{15}}}\:{arctan}\left(\frac{\mathrm{4}{t}+\mathrm{1}}{\sqrt{\mathrm{15}}}\right)\:+{c}_{\mathrm{2}} \:\Rightarrow{A}=\frac{\mathrm{2}}{\mathrm{5}}\:\frac{\mathrm{1}}{{tan}\left(\frac{{x}}{\mathrm{2}}\right)−\mathrm{1}}\:−\frac{\mathrm{1}}{\mathrm{5}\sqrt{\mathrm{15}}}\:{arctan}\left(\frac{\mathrm{4}{tan}\left(\frac{{x}}{\mathrm{2}}\right)+\mathrm{1}}{\sqrt{\mathrm{15}}}\right)\:+{C} \\ $$$$\Rightarrow{A}\:=\frac{\mathrm{2}}{\mathrm{5}}\:\:,\:{B}\:=−\frac{\mathrm{1}}{\mathrm{5}\sqrt{\mathrm{15}}}\:\:,\:{f}\left({x}\right)=\frac{\mathrm{4}{arctan}\left(\frac{{x}}{\mathrm{2}}\right)+\mathrm{1}}{\sqrt{\mathrm{15}}}\:. \\ $$$$ \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 21/Jan/19

(1/5)∫(((sinx+4)−(sinx−1))/((sinx+4)(sinx−1)))dx (1/5)∫(dx/(sinx−1))−(1/5)∫(dx/(sinx+4)) ∫(dx/(a+sinx)) ∫(dx/(a+((2tan(x/2))/(1+tan^2 (x/2))))) ∫((sec^2 (x/2)dx)/(a+atan^2 (x/2)+2tan(x/2))) k=tan(x/2) 2dk=sec^2 (x/2)dx ∫((2dk)/(a+ak^2 +2k)) (2/a)∫(dk/(1+k^2 +2(k/a))) (2/a)∫(dk/(k^2 +2×k×(1/a)+(1/a^2 )+1−(1/a^2 ))) (2/a)∫(dk/((k+(1/a))^2 +(((√(a^2 −1))/a))^2 )) (2/a)×(1/((((√(a^2 −1))/a))))×tan^(−1) (((k+(1/a))/((√(a^2 −1))/a)))+c =(2/(√(a^2 −1)))tan^(−1) (((ak+1)/(√(a^2 −1))))+c =(2/(√(a^2 −1)))tan^(−1) (((atan(x/2)+1)/(√(a^2 −1))))+c now (1/5)∫(dx/(−1+sinx))−(1/5)∫(dx/(4+sinx)) =(−(1/5))[∫(dx/(1−sinx))+∫(dx/(4+sinx))] =(((−1)/5))[∫((1+sinx)/(cos^2 x))dx+(2/(√(16−1)))tan^(−1) (((4tan(x/2)+1)/(√(4^2 −1))))] =(((−1)/5))[tanx+secx+(2/(√(15)))tan^(−1) (((4tan(x/2)+1)/(√(15))))] =(((−1)/5))[((((2tan(x/2))/(1+tan^2 (x/2)))+1)/((1−tan^2 (x/2))/(1+tan^2 (x/2))))+do] =(((−1)/5))[(((1+tan(x/2))^2 )/((1+tan(x/2))(1−tan(x/2))))+do] =(((−1)/5))(((1+tan(x/2))/(1−tan(x/2))))+(((−2)/(5(√(15)) )))(tan^(−1) (((4tan(x/2)+1)/((√(15)) )))

$$\frac{\mathrm{1}}{\mathrm{5}}\int\frac{\left({sinx}+\mathrm{4}\right)−\left({sinx}−\mathrm{1}\right)}{\left({sinx}+\mathrm{4}\right)\left({sinx}−\mathrm{1}\right)}{dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{5}}\int\frac{{dx}}{{sinx}−\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{5}}\int\frac{{dx}}{{sinx}+\mathrm{4}} \\ $$$$ \\ $$$$\int\frac{{dx}}{{a}+{sinx}} \\ $$$$\int\frac{{dx}}{{a}+\frac{\mathrm{2}{tan}\frac{{x}}{\mathrm{2}}}{\mathrm{1}+{tan}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}} \\ $$$$\int\frac{{sec}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}{dx}}{{a}+{atan}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}+\mathrm{2}{tan}\frac{{x}}{\mathrm{2}}} \\ $$$${k}={tan}\frac{{x}}{\mathrm{2}}\:\:\mathrm{2}{dk}={sec}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}{dx} \\ $$$$\int\frac{\mathrm{2}{dk}}{{a}+{ak}^{\mathrm{2}} +\mathrm{2}{k}} \\ $$$$\frac{\mathrm{2}}{{a}}\int\frac{{dk}}{\mathrm{1}+{k}^{\mathrm{2}} +\mathrm{2}\frac{{k}}{{a}}} \\ $$$$\frac{\mathrm{2}}{{a}}\int\frac{{dk}}{{k}^{\mathrm{2}} +\mathrm{2}×{k}×\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\mathrm{1}−\frac{\mathrm{1}}{{a}^{\mathrm{2}} }} \\ $$$$\frac{\mathrm{2}}{{a}}\int\frac{{dk}}{\left({k}+\frac{\mathrm{1}}{{a}}\right)^{\mathrm{2}} +\left(\frac{\sqrt{{a}^{\mathrm{2}} −\mathrm{1}}}{{a}}\right)^{\mathrm{2}} } \\ $$$$\frac{\mathrm{2}}{{a}}×\frac{\mathrm{1}}{\left(\frac{\sqrt{{a}^{\mathrm{2}} −\mathrm{1}}}{{a}}\right)}×{tan}^{−\mathrm{1}} \left(\frac{{k}+\frac{\mathrm{1}}{{a}}}{\frac{\sqrt{{a}^{\mathrm{2}} −\mathrm{1}}}{{a}}}\right)+{c} \\ $$$$=\frac{\mathrm{2}}{\sqrt{{a}^{\mathrm{2}} −\mathrm{1}}}{tan}^{−\mathrm{1}} \left(\frac{{ak}+\mathrm{1}}{\sqrt{{a}^{\mathrm{2}} −\mathrm{1}}}\right)+{c} \\ $$$$=\frac{\mathrm{2}}{\sqrt{{a}^{\mathrm{2}} −\mathrm{1}}}{tan}^{−\mathrm{1}} \left(\frac{{atan}\frac{{x}}{\mathrm{2}}+\mathrm{1}}{\sqrt{{a}^{\mathrm{2}} −\mathrm{1}}}\right)+{c} \\ $$$$ \\ $$$${now}\:\frac{\mathrm{1}}{\mathrm{5}}\int\frac{{dx}}{−\mathrm{1}+{sinx}}−\frac{\mathrm{1}}{\mathrm{5}}\int\frac{{dx}}{\mathrm{4}+{sinx}} \\ $$$$=\left(−\frac{\mathrm{1}}{\mathrm{5}}\right)\left[\int\frac{{dx}}{\mathrm{1}−{sinx}}+\int\frac{{dx}}{\mathrm{4}+{sinx}}\right] \\ $$$$=\left(\frac{−\mathrm{1}}{\mathrm{5}}\right)\left[\int\frac{\mathrm{1}+{sinx}}{{cos}^{\mathrm{2}} {x}}{dx}+\frac{\mathrm{2}}{\sqrt{\mathrm{16}−\mathrm{1}}}{tan}^{−\mathrm{1}} \left(\frac{\mathrm{4}{tan}\frac{{x}}{\mathrm{2}}+\mathrm{1}}{\sqrt{\mathrm{4}^{\mathrm{2}} −\mathrm{1}}}\right)\right] \\ $$$$=\left(\frac{−\mathrm{1}}{\mathrm{5}}\right)\left[{tanx}+{secx}+\frac{\mathrm{2}}{\sqrt{\mathrm{15}}}{tan}^{−\mathrm{1}} \left(\frac{\mathrm{4}{tan}\frac{{x}}{\mathrm{2}}+\mathrm{1}}{\sqrt{\mathrm{15}}}\right)\right] \\ $$$$=\left(\frac{−\mathrm{1}}{\mathrm{5}}\right)\left[\frac{\frac{\mathrm{2}{tan}\frac{{x}}{\mathrm{2}}}{\mathrm{1}+{tan}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}+\mathrm{1}}{\frac{\mathrm{1}−{tan}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}{\mathrm{1}+{tan}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}}+{do}\right] \\ $$$$=\left(\frac{−\mathrm{1}}{\mathrm{5}}\right)\left[\frac{\left(\mathrm{1}+{tan}\frac{{x}}{\mathrm{2}}\right)^{\mathrm{2}} }{\left(\mathrm{1}+{tan}\frac{{x}}{\mathrm{2}}\right)\left(\mathrm{1}−{tan}\frac{{x}}{\mathrm{2}}\right)}+{do}\right] \\ $$$$=\left(\frac{−\mathrm{1}}{\mathrm{5}}\right)\left(\frac{\mathrm{1}+{tan}\frac{{x}}{\mathrm{2}}}{\mathrm{1}−{tan}\frac{{x}}{\mathrm{2}}}\right)+\left(\frac{−\mathrm{2}}{\mathrm{5}\sqrt{\mathrm{15}}\:}\right)\left({tan}^{−\mathrm{1}} \left(\frac{\mathrm{4}{tan}\frac{{x}}{\mathrm{2}}+\mathrm{1}}{\sqrt{\mathrm{15}}\:}\right)\right. \\ $$

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