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1-2cosx-5sinx-3-dx-

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Question Number 71239 by 20190927 last updated on 13/Oct/19
∫(1/(2cosx−5sinx−3))dx
$$\int\frac{\mathrm{1}}{\mathrm{2cosx}−\mathrm{5sinx}−\mathrm{3}}\mathrm{dx} \\ $$
Commented by mathmax by abdo last updated on 13/Oct/19
let I =∫ (dx/(2cosx−5sinx −3)) changement tan((x/2))=t give I =∫ (1/(2((1−t^2 )/(1+t^2 ))−5((2t)/(1+t^2 ))−3))((2dt)/(1+t^2 )) =∫((2dt)/(2−2t^2 −10t−3−3t^2 )) =∫ ((2dt)/(−5t^2 −10t−1)) =∫((−2dt)/(5t^2 +10t +1)) 5t^2 +10t +1=0→Δ^′ =5^2 −5 =25−5=20 ⇒ t_1 =((−5+2(√5))/5) =−1+(2/( (√5))) and t_2 =−1−(2/( (√5))) ⇒ ∫ (dt/(5t^2 +10t+1)) =∫ (dt/(5(t−t_1 )(t−t_2 ))) =(1/(5(t_1 −t_2 )))∫ ((1/(t−t_1 ))−(1/(t−t_2 )))dt =(1/(5(4/( (√5))))) ∫ ((1/(t−t_1 ))−(1/(t−t_2 )))dt =(1/(4(√5))){ln∣t−t_1 ∣−ln∣t−t_2 ∣} +c ⇒I=−(1/(2(√5))){ln∣t−t_1 ∣−ln∣t−t_2 ∣} +c =−(1/(2(√5)))ln∣tan((x/2))+1−(2/( (√5)))∣+(1/(2(√5)))ln∣tan((x/2))+1+(2/( (√5)))∣ +C
$${let}\:{I}\:=\int\:\:\:\:\frac{{dx}}{\mathrm{2}{cosx}−\mathrm{5}{sinx}\:−\mathrm{3}}\:\:{changement}\:{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}\:{give} \\ $$$${I}\:=\int\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }−\mathrm{5}\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }−\mathrm{3}}\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }\:=\int\frac{\mathrm{2}{dt}}{\mathrm{2}−\mathrm{2}{t}^{\mathrm{2}} −\mathrm{10}{t}−\mathrm{3}−\mathrm{3}{t}^{\mathrm{2}} } \\ $$$$=\int\:\:\frac{\mathrm{2}{dt}}{−\mathrm{5}{t}^{\mathrm{2}} −\mathrm{10}{t}−\mathrm{1}}\:=\int\frac{−\mathrm{2}{dt}}{\mathrm{5}{t}^{\mathrm{2}} +\mathrm{10}{t}\:+\mathrm{1}} \\ $$$$\mathrm{5}{t}^{\mathrm{2}} \:+\mathrm{10}{t}\:+\mathrm{1}=\mathrm{0}\rightarrow\Delta^{'} =\mathrm{5}^{\mathrm{2}} −\mathrm{5}\:=\mathrm{25}−\mathrm{5}=\mathrm{20}\:\Rightarrow \\ $$$${t}_{\mathrm{1}} =\frac{−\mathrm{5}+\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{5}}\:=−\mathrm{1}+\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}\:\:{and}\:{t}_{\mathrm{2}} =−\mathrm{1}−\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}\:\Rightarrow \\ $$$$\int\:\:\frac{{dt}}{\mathrm{5}{t}^{\mathrm{2}} +\mathrm{10}{t}+\mathrm{1}}\:=\int\:\frac{{dt}}{\mathrm{5}\left({t}−{t}_{\mathrm{1}} \right)\left({t}−{t}_{\mathrm{2}} \right)}\:=\frac{\mathrm{1}}{\mathrm{5}\left({t}_{\mathrm{1}} −{t}_{\mathrm{2}} \right)}\int\:\:\left(\frac{\mathrm{1}}{{t}−{t}_{\mathrm{1}} }−\frac{\mathrm{1}}{{t}−{t}_{\mathrm{2}} }\right){dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{5}\frac{\mathrm{4}}{\:\sqrt{\mathrm{5}}}}\:\int\:\left(\frac{\mathrm{1}}{{t}−{t}_{\mathrm{1}} }−\frac{\mathrm{1}}{{t}−{t}_{\mathrm{2}} }\right){dt}\:=\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{5}}}\left\{{ln}\mid{t}−{t}_{\mathrm{1}} \mid−{ln}\mid{t}−{t}_{\mathrm{2}} \mid\right\}\:+{c} \\ $$$$\Rightarrow{I}=−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{5}}}\left\{{ln}\mid{t}−{t}_{\mathrm{1}} \mid−{ln}\mid{t}−{t}_{\mathrm{2}} \mid\right\}\:+{c} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{5}}}{ln}\mid{tan}\left(\frac{{x}}{\mathrm{2}}\right)+\mathrm{1}−\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}\mid+\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{5}}}{ln}\mid{tan}\left(\frac{{x}}{\mathrm{2}}\right)+\mathrm{1}+\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}\mid\:+{C} \\ $$
Commented by 20190927 last updated on 13/Oct/19
thank you very much
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much} \\ $$
Commented by turbo msup by abdo last updated on 13/Oct/19
you are welcome
$${you}\:{are}\:{welcome} \\ $$
Answered by MJS last updated on 13/Oct/19
∫(dx/(2cos x −5sin x −3))= [t=tan (x/2) → dx=((2dt)/(t^2 +1))] =−2∫ (dt/(5t^2 +10t+1))= =((√5)/2)∫(dt/(5t+5+2(√5)))−((√5)/2)∫(dt/(5t+5−2(√5)))= =((√5)/(10))ln ((5t+5+2(√5))/(5t+5−2(√5))) =...
$$\int\frac{{dx}}{\mathrm{2cos}\:{x}\:−\mathrm{5sin}\:{x}\:−\mathrm{3}}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\:\rightarrow\:{dx}=\frac{\mathrm{2}{dt}}{{t}^{\mathrm{2}} +\mathrm{1}}\right] \\ $$$$=−\mathrm{2}\int\:\frac{{dt}}{\mathrm{5}{t}^{\mathrm{2}} +\mathrm{10}{t}+\mathrm{1}}= \\ $$$$=\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\int\frac{{dt}}{\mathrm{5}{t}+\mathrm{5}+\mathrm{2}\sqrt{\mathrm{5}}}−\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\int\frac{{dt}}{\mathrm{5}{t}+\mathrm{5}−\mathrm{2}\sqrt{\mathrm{5}}}= \\ $$$$=\frac{\sqrt{\mathrm{5}}}{\mathrm{10}}\mathrm{ln}\:\frac{\mathrm{5}{t}+\mathrm{5}+\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{5}{t}+\mathrm{5}−\mathrm{2}\sqrt{\mathrm{5}}}\:=… \\ $$
Answered by peter frank last updated on 13/Oct/19
t=tan(x/2) dx=(2/(1+t^2 ))dt 2(((1−t^2 )/(1+t^2 )))−5(((2t)/(1+t^2 )))−3 ((2−2t^2 −10t−3(1+t^2 ))/(1+t^2 )) ((2−2t^2 −10t−3−3t^2 )/(1+t^2 )) ((−5t^2 −10t−5)/(1+t^2 )) ∫((2/(1+t^2 ))/((−5t^2 −10t−5)/(1+t^2 )))dt ∫((2dt)/(t^2 +2t+1)) ∫((2dt)/((t+1)^2 −((2/( (√5))))^2 )) ........
$${t}=\mathrm{tan}\frac{{x}}{\mathrm{2}}\: \\ $$$${dx}=\frac{\mathrm{2}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$$\mathrm{2}\left(\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }\right)−\mathrm{5}\left(\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }\right)−\mathrm{3} \\ $$$$\frac{\mathrm{2}−\mathrm{2}{t}^{\mathrm{2}} −\mathrm{10}{t}−\mathrm{3}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$\frac{\mathrm{2}−\mathrm{2}{t}^{\mathrm{2}} −\mathrm{10}{t}−\mathrm{3}−\mathrm{3}{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$\frac{−\mathrm{5}{t}^{\mathrm{2}} −\mathrm{10}{t}−\mathrm{5}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$\int\frac{\frac{\mathrm{2}}{\mathrm{1}+{t}^{\mathrm{2}} }}{\frac{−\mathrm{5}{t}^{\mathrm{2}} −\mathrm{10}{t}−\mathrm{5}}{\mathrm{1}+{t}^{\mathrm{2}} }}{dt} \\ $$$$\int\frac{\mathrm{2}{dt}}{{t}^{\mathrm{2}} +\mathrm{2}{t}+\mathrm{1}} \\ $$$$\int\frac{\mathrm{2}{dt}}{\left({t}+\mathrm{1}\right)^{\mathrm{2}} −\left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}\right)^{\mathrm{2}} } \\ $$$$…….. \\ $$

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