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Question-84681

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Question Number 84681 by Power last updated on 15/Mar/20
Commented by mathmax by abdo last updated on 15/Mar/20
I =∫ ((15sinx +2cosx)/(98sinx −7cosx))dx ⇒I =((15)/(98))∫ ((sinx+(2/(15))cosx)/(sinx−(7/(98))cosx))dx let determine A =∫ ((sinx+acosx)/(sinx +bcosx))dx we di the changement tan((x/2))=t ⇒ A =∫((((2t)/(1+t^2 ))+a((1−t^2 )/(1+t^2 )))/(((2t)/(1+t^2 ))+b((1−t^2 )/(1+t^2 )))) ×((2dt)/(1+t^2 )) =2 ∫ ((2t+a−at^2 )/((2t+b−bt^2 )(1+t^2 )))dt =2 ∫ ((−at^2 +2t+a)/((−bt^2 +2t+b)(t^2 +1)))dt =2∫ ((at^2 −2t−a)/((t^2 +1)(bt^2 −2t−b)))dt let decompose F(t) =((at^2 −2t−a)/((t^2 +1)(bt^2 −2t −b))) bt^2 −2t−b=0→Δ^′ =1+b^2 >0 ⇒t_1 =((1+(√(b^2 +1)))/b) t_2 =((1−(√(b^2 +1)))/b) ⇒F(t) =((at^2 −2t−a)/(b(t−t_1 )(t−t_2 )(t^2 +1))) =(α/(t−t_1 )) +(β/(t−t_2 )) +((λt +γ)/(t^2 +1)) α =((at_1 ^2 −2t_1 −a)/(b(t_1 −t_2 )(t_1 ^2 +1))) β =((at_2 ^2 −2t_2 −a)/(b(t_2 −t_1 )(t_2 ^2 +1))) ....be continued...
$${I}\:=\int\:\:\frac{\mathrm{15}{sinx}\:+\mathrm{2}{cosx}}{\mathrm{98}{sinx}\:−\mathrm{7}{cosx}}{dx}\:\Rightarrow{I}\:=\frac{\mathrm{15}}{\mathrm{98}}\int\:\:\frac{{sinx}+\frac{\mathrm{2}}{\mathrm{15}}{cosx}}{{sinx}−\frac{\mathrm{7}}{\mathrm{98}}{cosx}}{dx} \\ $$$${let}\:{determine}\:{A}\:=\int\:\:\frac{{sinx}+{acosx}}{{sinx}\:+{bcosx}}{dx}\:\:{we}\:{di}\:{the}\:{changement} \\ $$$${tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}\:\Rightarrow\:{A}\:=\int\frac{\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }+{a}\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}{\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }+{b}\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}\:×\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$=\mathrm{2}\:\int\:\:\frac{\mathrm{2}{t}+{a}−{at}^{\mathrm{2}} }{\left(\mathrm{2}{t}+{b}−{bt}^{\mathrm{2}} \right)\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{dt}\:=\mathrm{2}\:\int\:\:\frac{−{at}^{\mathrm{2}} +\mathrm{2}{t}+{a}}{\left(−{bt}^{\mathrm{2}} +\mathrm{2}{t}+{b}\right)\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)}{dt} \\ $$$$=\mathrm{2}\int\:\:\frac{{at}^{\mathrm{2}} −\mathrm{2}{t}−{a}}{\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)\left({bt}^{\mathrm{2}} −\mathrm{2}{t}−{b}\right)}{dt}\:\:{let}\:{decompose} \\ $$$${F}\left({t}\right)\:=\frac{{at}^{\mathrm{2}} −\mathrm{2}{t}−{a}}{\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)\left({bt}^{\mathrm{2}} −\mathrm{2}{t}\:−{b}\right)} \\ $$$${bt}^{\mathrm{2}} −\mathrm{2}{t}−{b}=\mathrm{0}\rightarrow\Delta^{'} =\mathrm{1}+{b}^{\mathrm{2}} \:>\mathrm{0}\:\Rightarrow{t}_{\mathrm{1}} =\frac{\mathrm{1}+\sqrt{{b}^{\mathrm{2}} +\mathrm{1}}}{{b}} \\ $$$${t}_{\mathrm{2}} =\frac{\mathrm{1}−\sqrt{{b}^{\mathrm{2}} \:+\mathrm{1}}}{{b}}\:\Rightarrow{F}\left({t}\right)\:=\frac{{at}^{\mathrm{2}} −\mathrm{2}{t}−{a}}{{b}\left({t}−{t}_{\mathrm{1}} \right)\left({t}−{t}_{\mathrm{2}} \right)\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)} \\ $$$$=\frac{\alpha}{{t}−{t}_{\mathrm{1}} }\:+\frac{\beta}{{t}−{t}_{\mathrm{2}} }\:+\frac{\lambda{t}\:+\gamma}{{t}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$$\alpha\:=\frac{{at}_{\mathrm{1}} ^{\mathrm{2}} −\mathrm{2}{t}_{\mathrm{1}} −{a}}{{b}\left({t}_{\mathrm{1}} −{t}_{\mathrm{2}} \right)\left({t}_{\mathrm{1}} ^{\mathrm{2}} \:+\mathrm{1}\right)} \\ $$$$\beta\:=\frac{{at}_{\mathrm{2}} ^{\mathrm{2}} −\mathrm{2}{t}_{\mathrm{2}} −{a}}{{b}\left({t}_{\mathrm{2}} −{t}_{\mathrm{1}} \right)\left({t}_{\mathrm{2}} ^{\mathrm{2}} \:+\mathrm{1}\right)}\:\:….{be}\:{continued}… \\ $$$$ \\ $$
Answered by TANMAY PANACEA last updated on 15/Mar/20
∫((asinx+bcosx)/(csinx+dcosx))dx asinx+bcosx=p(csinx+dcosx)+q×(d/dx) (csinx+dcosx) so ∫((p(csinx+dcosx)+q×(d/dx)(csinx+dcosx))/(csinx+dcosx))dx ∫pdx+q∫((d(csinx+dcosx))/(csinx+dcosx)) px+qln(csinx+dcosx)+C now 15sinx+2cosx=p(98sinx−7cosx)+q(98cosx+7sinx) 98p+7q=15 −7p+98q=2 ×14 98p+7q=15 −98p+98×14q=28 q×7(1+196)=28 q=((28)/(7×197))=(4/(197)) 98p=15−7×(4/(197))=((15×197−28)/(197)) p=(((15×197−28)/(197×98))) pls calculate p and q...
$$\int\frac{{asinx}+{bcosx}}{{csinx}+{dcosx}}{dx} \\ $$$${asinx}+{bcosx}={p}\left({csinx}+{dcosx}\right)+{q}×\frac{{d}}{{dx}}\:\left({csinx}+{dcosx}\right) \\ $$$${so} \\ $$$$\int\frac{{p}\left({csinx}+{dcosx}\right)+{q}×\frac{{d}}{{dx}}\left({csinx}+{dcosx}\right)}{{csinx}+{dcosx}}{dx} \\ $$$$\int{pdx}+{q}\int\frac{{d}\left({csinx}+{dcosx}\right)}{{csinx}+{dcosx}} \\ $$$${px}+{qln}\left({csinx}+{dcosx}\right)+{C} \\ $$$${now}\: \\ $$$$\mathrm{15}{sinx}+\mathrm{2}{cosx}={p}\left(\mathrm{98}{sinx}−\mathrm{7}{cosx}\right)+{q}\left(\mathrm{98}{cosx}+\mathrm{7}{sinx}\right) \\ $$$$\mathrm{98}{p}+\mathrm{7}{q}=\mathrm{15} \\ $$$$−\mathrm{7}{p}+\mathrm{98}{q}=\mathrm{2}\:\:×\mathrm{14} \\ $$$$\mathrm{98}{p}+\mathrm{7}{q}=\mathrm{15} \\ $$$$−\mathrm{98}{p}+\mathrm{98}×\mathrm{14}{q}=\mathrm{28} \\ $$$${q}×\mathrm{7}\left(\mathrm{1}+\mathrm{196}\right)=\mathrm{28}\:\:\boldsymbol{{q}}=\frac{\mathrm{28}}{\mathrm{7}×\mathrm{197}}=\frac{\mathrm{4}}{\mathrm{197}} \\ $$$$\mathrm{98}\boldsymbol{{p}}=\mathrm{15}−\mathrm{7}×\frac{\mathrm{4}}{\mathrm{197}}=\frac{\mathrm{15}×\mathrm{197}−\mathrm{28}}{\mathrm{197}} \\ $$$$\boldsymbol{{p}}=\left(\frac{\mathrm{15}×\mathrm{197}−\mathrm{28}}{\mathrm{197}×\mathrm{98}}\right) \\ $$$$\boldsymbol{{pls}}\:\boldsymbol{{calculate}}\:\boldsymbol{{p}}\:\boldsymbol{{and}}\:\boldsymbol{{q}}… \\ $$
Commented by Power last updated on 15/Mar/20
thanks
$$\mathrm{thanks} \\ $$

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