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Question Number 116999 by TANMAY PANACEA last updated on 08/Oct/20

∫(dx/((a+bcosx)^2 ))

$$\int\frac{{dx}}{\left({a}+{bcosx}\right)^{\mathrm{2}} } \\ $$

Answered by Olaf last updated on 08/Oct/20

I(x) = ∫(dx/((a+b((e^(ix) +e^(−ix) )/2))^2 )) u = e^(ix) du = ie^(ix) dx = iudx dx = −i(du/u) I(u) = −i∫((du/u)/((a+b((u+(1/u))/2))^2 )) I(u) = −i∫((udu)/((au+b((u^2 +1)/2))^2 )) I(u) = −i∫((4udu)/((bu^2 +2au+b)^2 )) I(u) = −((4i)/b^2 )∫((4udu)/([(u+(a/b))^2 +1−(a^2 /b^2 )]^2 )) Different cases in function of sign of 1−(a^2 /b^2 )...

$$\mathrm{I}\left({x}\right)\:=\:\int\frac{{dx}}{\left({a}+{b}\frac{{e}^{{ix}} +{e}^{−{ix}} }{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$${u}\:=\:{e}^{{ix}} \\ $$$${du}\:=\:{ie}^{{ix}} {dx}\:=\:{iudx} \\ $$$${dx}\:=\:−{i}\frac{{du}}{{u}} \\ $$$$\mathrm{I}\left({u}\right)\:=\:−{i}\int\frac{\frac{{du}}{{u}}}{\left({a}+{b}\frac{{u}+\frac{\mathrm{1}}{{u}}}{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$$\mathrm{I}\left({u}\right)\:=\:−{i}\int\frac{{udu}}{\left({au}+{b}\frac{{u}^{\mathrm{2}} +\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$$\mathrm{I}\left({u}\right)\:=\:−{i}\int\frac{\mathrm{4}{udu}}{\left({bu}^{\mathrm{2}} +\mathrm{2}{au}+{b}\right)^{\mathrm{2}} } \\ $$$$\mathrm{I}\left({u}\right)\:=\:−\frac{\mathrm{4}{i}}{{b}^{\mathrm{2}} }\int\frac{\mathrm{4}{udu}}{\left[\left({u}+\frac{{a}}{{b}}\right)^{\mathrm{2}} +\mathrm{1}−\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\right]^{\mathrm{2}} } \\ $$$$\mathrm{Different}\:\mathrm{cases}\:\mathrm{in}\:\mathrm{function}\:\mathrm{of}\:\mathrm{sign}\:\mathrm{of} \\ $$$$\mathrm{1}−\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }... \\ $$

Answered by mathmax by abdo last updated on 08/Oct/20

let ϕ(a) =∫ (dx/(a+bcosx)) ⇒ϕ^′ (a) =−∫ (dx/((a+bcosx)^2 )) ⇒ ∫ (dx/((a+bcosx)^2 ))=−ϕ^′ (a) but ϕ(a) =_(tan((x/2))=t) ∫ ((2dt)/((1+t^2 )(a+b((1−t^2 )/(1+t^2 ))))) =∫ ((2dt)/(a+at^2 +b−bt^2 )) =2∫ (dt/((a−b)t^2 +a+b)) =(2/(a−b)) ∫ (dt/(t^2 +((a+b)/(a−b)))) case 1 ((a+b)/(a−b))>0 we do ch.t =(√((a+b)/(a−b)))z ⇒ ϕ(a) =(2/(a−b)) ×((a−b)/(a+b)) ∫ (1/(1+z^2 ))((√(a+b))/(√(a−b)))dz =(2/(√(a^2 −b^2 ))) arctan((√((a−b)/(a+b)))t) =(2/(√(a^2 −b^2 ))) arctan((√((a−b)/(a+b)))tan((x/2)))+c ⇒ ϕ^′ (a) =2(a^2 −b^2 )^(−(1/2)) =−2a(a^2 −b^2 )^(−(3/2)) arctan((√((a−b)/(a+b)))tan((x/2))) +(2/(√(a^2 −b^2 )))tan((x/2))×(((((a−b)/(a+b)))^′ )/(2(√((a−b)/(a+b)))(1+((a−b)/(a+b))tan^2 ((x/2))))) with (((a−b)/(a+b)))^′ =((a+b−(a−b))/((a+b)^2 )) =((2b)/((a+b)^2 )) case 2 ((a+b)/(a−b))<0 ⇒ϕ(a) =(2/(a−b))∫ (dt/(t^2 −(((a+b)/(b−a))))) =_(t =(√((b+a)/(b−a)))u) (2/(a−b)) ×((b−a)/(b+a)) ∫ (1/(u^2 −1))×((√(b+a))/(√(b−a))) du =((−2)/(√(b^2 −a^2 )))∫((1/(u−1))−(1/(u+1)))du =((−2)/(√(b^2 −a^2 )))ln∣((u−1)/(u+1))∣ +c =((−2)/(√(b^2 −a^2 )))ln∣(((√((b−a)/(b+a)))tan((x/2))−1)/((√((b−a)/(b+a)))tan((x/2))+1))∣ +c rest calculus of ϕ^′ (a)...

$$\mathrm{let}\:\:\varphi\left(\mathrm{a}\right)\:=\int\:\:\frac{\mathrm{dx}}{\mathrm{a}+\mathrm{bcosx}}\:\Rightarrow\varphi^{'} \left(\mathrm{a}\right)\:=−\int\:\:\frac{\mathrm{dx}}{\left(\mathrm{a}+\mathrm{bcosx}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$$\int\:\frac{\mathrm{dx}}{\left(\mathrm{a}+\mathrm{bcosx}\right)^{\mathrm{2}} }=−\varphi^{'} \left(\mathrm{a}\right)\:\:\mathrm{but}\:\varphi\left(\mathrm{a}\right)\:=_{\mathrm{tan}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)=\mathrm{t}} \:\:\int\:\:\:\frac{\mathrm{2dt}}{\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)\left(\mathrm{a}+\mathrm{b}\frac{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\right)} \\ $$$$=\int\:\:\frac{\mathrm{2dt}}{\mathrm{a}+\mathrm{at}^{\mathrm{2}} \:+\mathrm{b}−\mathrm{bt}^{\mathrm{2}} }\:=\mathrm{2}\int\:\:\frac{\mathrm{dt}}{\left(\mathrm{a}−\mathrm{b}\right)\mathrm{t}^{\mathrm{2}} \:+\mathrm{a}+\mathrm{b}} \\ $$$$=\frac{\mathrm{2}}{\mathrm{a}−\mathrm{b}}\:\int\:\:\frac{\mathrm{dt}}{\mathrm{t}^{\mathrm{2}} +\frac{\mathrm{a}+\mathrm{b}}{\mathrm{a}−\mathrm{b}}}\:\:\mathrm{case}\:\mathrm{1}\:\:\frac{\mathrm{a}+\mathrm{b}}{\mathrm{a}−\mathrm{b}}>\mathrm{0}\:\:\mathrm{we}\:\mathrm{do}\:\mathrm{ch}.\mathrm{t}\:=\sqrt{\frac{\mathrm{a}+\mathrm{b}}{\mathrm{a}−\mathrm{b}}}\mathrm{z}\:\Rightarrow \\ $$$$\varphi\left(\mathrm{a}\right)\:=\frac{\mathrm{2}}{\mathrm{a}−\mathrm{b}}\:×\frac{\mathrm{a}−\mathrm{b}}{\mathrm{a}+\mathrm{b}}\:\int\:\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{z}^{\mathrm{2}} }\frac{\sqrt{\mathrm{a}+\mathrm{b}}}{\sqrt{\mathrm{a}−\mathrm{b}}}\mathrm{dz}\:=\frac{\mathrm{2}}{\sqrt{\mathrm{a}^{\mathrm{2}} −\mathrm{b}^{\mathrm{2}} }}\:\mathrm{arctan}\left(\sqrt{\frac{\mathrm{a}−\mathrm{b}}{\mathrm{a}+\mathrm{b}}}\mathrm{t}\right) \\ $$$$=\frac{\mathrm{2}}{\sqrt{\mathrm{a}^{\mathrm{2}} −\mathrm{b}^{\mathrm{2}} }}\:\mathrm{arctan}\left(\sqrt{\frac{\mathrm{a}−\mathrm{b}}{\mathrm{a}+\mathrm{b}}}\mathrm{tan}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)\right)+\mathrm{c}\:\Rightarrow \\ $$$$\varphi^{'} \left(\mathrm{a}\right)\:=\mathrm{2}\left(\mathrm{a}^{\mathrm{2}} −\mathrm{b}^{\mathrm{2}} \right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \:=−\mathrm{2a}\left(\mathrm{a}^{\mathrm{2}} −\mathrm{b}^{\mathrm{2}} \right)^{−\frac{\mathrm{3}}{\mathrm{2}}} \:\mathrm{arctan}\left(\sqrt{\frac{\mathrm{a}−\mathrm{b}}{\mathrm{a}+\mathrm{b}}}\mathrm{tan}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)\right) \\ $$$$+\frac{\mathrm{2}}{\sqrt{\mathrm{a}^{\mathrm{2}} −\mathrm{b}^{\mathrm{2}} }}\mathrm{tan}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)×\frac{\left(\frac{\mathrm{a}−\mathrm{b}}{\mathrm{a}+\mathrm{b}}\right)^{'} }{\mathrm{2}\sqrt{\frac{\mathrm{a}−\mathrm{b}}{\mathrm{a}+\mathrm{b}}}\left(\mathrm{1}+\frac{\mathrm{a}−\mathrm{b}}{\mathrm{a}+\mathrm{b}}\mathrm{tan}^{\mathrm{2}} \left(\frac{\mathrm{x}}{\mathrm{2}}\right)\right)}\:\mathrm{with} \\ $$$$\left(\frac{\mathrm{a}−\mathrm{b}}{\mathrm{a}+\mathrm{b}}\right)^{'} \:=\frac{\mathrm{a}+\mathrm{b}−\left(\mathrm{a}−\mathrm{b}\right)}{\left(\mathrm{a}+\mathrm{b}\right)^{\mathrm{2}} }\:=\frac{\mathrm{2b}}{\left(\mathrm{a}+\mathrm{b}\right)^{\mathrm{2}} } \\ $$$$\mathrm{case}\:\mathrm{2}\:\:\frac{\mathrm{a}+\mathrm{b}}{\mathrm{a}−\mathrm{b}}<\mathrm{0}\:\Rightarrow\varphi\left(\mathrm{a}\right)\:=\frac{\mathrm{2}}{\mathrm{a}−\mathrm{b}}\int\:\:\frac{\mathrm{dt}}{\mathrm{t}^{\mathrm{2}} −\left(\frac{\mathrm{a}+\mathrm{b}}{\mathrm{b}−\mathrm{a}}\right)} \\ $$$$=_{\mathrm{t}\:=\sqrt{\frac{\mathrm{b}+\mathrm{a}}{\mathrm{b}−\mathrm{a}}}\mathrm{u}} \:\:\:\frac{\mathrm{2}}{\mathrm{a}−\mathrm{b}}\:×\frac{\mathrm{b}−\mathrm{a}}{\mathrm{b}+\mathrm{a}}\:\int\:\:\frac{\mathrm{1}}{\mathrm{u}^{\mathrm{2}} −\mathrm{1}}×\frac{\sqrt{\mathrm{b}+\mathrm{a}}}{\sqrt{\mathrm{b}−\mathrm{a}}}\:\mathrm{du} \\ $$$$=\frac{−\mathrm{2}}{\sqrt{\mathrm{b}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} }}\int\left(\frac{\mathrm{1}}{\mathrm{u}−\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{u}+\mathrm{1}}\right)\mathrm{du}\:=\frac{−\mathrm{2}}{\sqrt{\mathrm{b}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} }}\mathrm{ln}\mid\frac{\mathrm{u}−\mathrm{1}}{\mathrm{u}+\mathrm{1}}\mid\:+\mathrm{c} \\ $$$$=\frac{−\mathrm{2}}{\sqrt{\mathrm{b}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} }}\mathrm{ln}\mid\frac{\sqrt{\frac{\mathrm{b}−\mathrm{a}}{\mathrm{b}+\mathrm{a}}}\mathrm{tan}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)−\mathrm{1}}{\sqrt{\frac{\mathrm{b}−\mathrm{a}}{\mathrm{b}+\mathrm{a}}}\mathrm{tan}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)+\mathrm{1}}\mid\:+\mathrm{c} \\ $$$$\mathrm{rest}\:\mathrm{calculus}\:\mathrm{of}\:\varphi^{'} \left(\mathrm{a}\right)... \\ $$

Answered by TANMAY PANACEA last updated on 09/Oct/20

p=((sinx)/(a+bcosx))→(dp/dx)=(((a+bcosx)cosx−sinx(−bsinx))/((a+bcosx)^2 )) (dp/dx)=((acosx+b)/((a+bcosx)^2 ))=(((a/b)(bcosx+a−a)+b)/((a+bcosx)^2 )) (dp/dx)=((a/b)/((a+bcosx)))+(((b^2 −a^2 )/b)/((a+bcosx)^2 )) ∫d(((sinx)/(a+bcosx)))=(a/b)∫(dx/(b((a/b)+((1−tan^2 (x/2))/(1+tan^2 (x/2))))))+((b^2 −a^2 )/b)×I ((sinx)/(a+bcosx))=(a/b^2 )∫((sec^2 (x/2))/((a/b)+1+((a/b)−1)tan^2 (x/2)))dx+((b^2 −a^2 )/b)×I ((sinx)/(a+bcosx))=(a/b^2 )∫((sec^2 (x/2))/(((a/b)−1)(((a+b)/b)×(b/(a−b))+tan^2 (x/2))))+((b^2 −a^2 )/b)I ((sinx)/(a+bcosx))=(a/b^2 )×(b/((a−b)))∫((d(tan(x/2))×2)/({((√((a+b)/(a−b))) )^2 +tan^2 (x/2)))+((b^2 −a^2 )/b)×I ((b/(b^2 −a^2 )))(((sinx)/(a+bcosx)))=(a/b)×(1/(a−b))×((2b)/((a+b)(a−b)))×(1/( (√((a+b)/(a−b)))))×tan^(−1) (((tan(x/2))/( (√((a+b)/(a−b))))))+I so I= (b/(b^2 −a^2 ))(((sinx)/(a+bcosx)))−((2a)/((a+b)(a−b)^2 ))×((√(a−b))/( (√(a+b))))×tan^(−1) (((tan(x/2))/( (√((a+b)/(a−b))))))+C I=(b/(b^2 −a^2 ))(((sinx)/(a+bcosx)))−((2a)/((a^2 −b^2 )^(3/2) ))tan^(−1) (((tan(x/2))/( (√((a+b)/(a−b))))))+C pls check

$${p}=\frac{{sinx}}{{a}+{bcosx}}\rightarrow\frac{{dp}}{{dx}}=\frac{\left({a}+{bcosx}\right){cosx}−{sinx}\left(−{bsinx}\right)}{\left({a}+{bcosx}\right)^{\mathrm{2}} } \\ $$$$\frac{{dp}}{{dx}}=\frac{{acosx}+{b}}{\left({a}+{bcosx}\right)^{\mathrm{2}} }=\frac{\frac{{a}}{{b}}\left({bcosx}+{a}−{a}\right)+{b}}{\left({a}+{bcosx}\right)^{\mathrm{2}} } \\ $$$$\frac{{dp}}{{dx}}=\frac{\frac{{a}}{{b}}}{\left({a}+{bcosx}\right)}+\frac{\frac{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }{{b}}}{\left({a}+{bcosx}\right)^{\mathrm{2}} } \\ $$$$\int{d}\left(\frac{{sinx}}{{a}+{bcosx}}\right)=\frac{{a}}{{b}}\int\frac{{dx}}{{b}\left(\frac{{a}}{{b}}+\frac{\mathrm{1}−{tan}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}{\mathrm{1}+{tan}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}\right)}+\frac{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }{{b}}×{I} \\ $$$$\frac{{sinx}}{{a}+{bcosx}}=\frac{{a}}{{b}^{\mathrm{2}} }\int\frac{{sec}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}{\frac{{a}}{{b}}+\mathrm{1}+\left(\frac{{a}}{{b}}−\mathrm{1}\right){tan}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}{dx}+\frac{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }{{b}}×{I} \\ $$$$\frac{{sinx}}{{a}+{bcosx}}=\frac{{a}}{{b}^{\mathrm{2}} }\int\frac{{sec}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}{\left(\frac{{a}}{{b}}−\mathrm{1}\right)\left(\frac{{a}+{b}}{{b}}×\frac{{b}}{{a}−{b}}+{tan}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}\right)}+\frac{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }{{b}}\boldsymbol{{I}} \\ $$$$\frac{{sinx}}{{a}+{bcosx}}=\frac{{a}}{{b}^{\mathrm{2}} }×\frac{{b}}{\left({a}−{b}\right)}\int\frac{{d}\left({tan}\frac{{x}}{\mathrm{2}}\right)×\mathrm{2}}{\left\{\left(\sqrt{\frac{{a}+{b}}{{a}−{b}}}\:\right)^{\mathrm{2}} +{tan}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}\right.}+\frac{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }{{b}}×{I} \\ $$$$\left(\frac{{b}}{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }\right)\left(\frac{{sinx}}{{a}+{bcosx}}\right)=\frac{{a}}{{b}}×\frac{\mathrm{1}}{{a}−{b}}×\frac{\mathrm{2}{b}}{\left({a}+{b}\right)\left({a}−{b}\right)}×\frac{\mathrm{1}}{\:\sqrt{\frac{{a}+{b}}{{a}−{b}}}}×{tan}^{−\mathrm{1}} \left(\frac{{tan}\frac{{x}}{\mathrm{2}}}{\:\sqrt{\frac{{a}+{b}}{{a}−{b}}}}\right)+\boldsymbol{{I}} \\ $$$${so}\:{I}= \\ $$$$\frac{{b}}{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }\left(\frac{{sinx}}{{a}+{bcosx}}\right)−\frac{\mathrm{2}{a}}{\left({a}+{b}\right)\left({a}−{b}\right)^{\mathrm{2}} }×\frac{\sqrt{{a}−{b}}}{\:\sqrt{{a}+{b}}}×{tan}^{−\mathrm{1}} \left(\frac{{tan}\frac{{x}}{\mathrm{2}}}{\:\sqrt{\frac{{a}+{b}}{{a}−{b}}}}\right)+{C} \\ $$$${I}=\frac{{b}}{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }\left(\frac{{sinx}}{{a}+{bcosx}}\right)−\frac{\mathrm{2}{a}}{\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{tan}^{−\mathrm{1}} \left(\frac{{tan}\frac{{x}}{\mathrm{2}}}{\:\sqrt{\frac{{a}+{b}}{{a}−{b}}}}\right)+{C} \\ $$$$\boldsymbol{{pls}}\:\boldsymbol{{check}} \\ $$$$ \\ $$

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