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x-a-2-cosx-b-2-sinx-dx-

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Question Number 103537 by TMSF last updated on 15/Jul/20
∫(x/((a^2 cosx+b^2 sinx)))dx
$$\int\frac{\mathrm{x}}{\left(\mathrm{a}^{\mathrm{2}} \mathrm{cosx}+\mathrm{b}^{\mathrm{2}} \mathrm{sinx}\right)}\mathrm{dx} \\ $$
Answered by mathmax by abdo last updated on 15/Jul/20
let I =∫ ((xdx)/(a^2 cosx +b^2 sinx)) we put a^2 =α and b^2 =β ⇒ I =∫ ((xdx)/(αcosx +βsinx)) changement tan((x/2))=t give I =∫ ((2arctan(t))/(α×((1−t^2 )/(1+t^2 )) +β×((2t)/(1+t^2 ))))×((2dt)/(1+t^2 )) =4∫ ((arctan(t))/(α−αt^2 +2βt))dt =−4 ∫ ((arctan(t))/(αt^2 −2βt −α))dt let f(ξ) =∫ ((arctan(ξt))/(αt^2 −2βt −α))dt we have f^′ (ξ) =∫ (t/((1+ξ^2 t^2 )(αt^2 −2βt −α)))dt =_(ξt =u) ∫ (u/(ξ(1+u^2 )(α(u^2 /ξ^2 )−2β(u/ξ)−α)))(du/ξ) =∫ ((udu)/(ξ^2 (u^2 +1)(α (u^2 /ξ^2 ) −2β (u/ξ)−α))) =∫ ((udu)/((u^2 +1)(αu^2 −2βξu −αξ^2 ))) let decompose F(u) =(u/((u^2 +1)(αu^2 −2βξu −αξ^2 ))) αu^2 −2βξu −αξ^2 →Δ^′ =β^2 ξ^2 −α^2 ξ^2 =ξ^2 (β^2 −α^2 ) =ξ^2 (b^4 −a^4 ) case1 ∣b∣>∣a∣ ⇒u_1 =((βξ+ξ(√(β^2 −α^2 )))/α) and u_2 =((βξ−ξ(√(β^2 −α^2 )))/α) ⇒ F(u) =(u/((u^2 +1)α(u−u_1 )(u−u_2 ))) =(a_1 /(u−u_1 )) +(a_2 /(u−u_2 )) +((b_1 u +b_2 )/(u^2 +1)) eazy to find a_i and b_i ⇒ f^′ (ξ) =a_1 ln∣u−u_1 ∣ +a_2 ln∣u−u_2 ∣ +(b_1 /2)ln(u^2 +1)+b_2 arctan(u) +c ⇒ f(ξ) =a_1 ∫ ln∣u−u_1 ∣dξ +a_2 ∫ ln∣u−u_2 ∣dξ +(b_1 /2) ∫ ln∣u^2 +1∣dξ +b_2 ∫ arctanu dξ +cξ ...be continued...
$$\mathrm{let}\:\mathrm{I}\:=\int\:\:\frac{\mathrm{xdx}}{\mathrm{a}^{\mathrm{2}} \mathrm{cosx}\:+\mathrm{b}^{\mathrm{2}} \mathrm{sinx}}\:\:\mathrm{we}\:\mathrm{put}\:\mathrm{a}^{\mathrm{2}} \:=\alpha\:\mathrm{and}\:\mathrm{b}^{\mathrm{2}} \:=\beta\:\Rightarrow \\ $$$$\mathrm{I}\:=\int\:\:\frac{\mathrm{xdx}}{\alpha\mathrm{cosx}\:+\beta\mathrm{sinx}}\:\mathrm{changement}\:\mathrm{tan}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)=\mathrm{t}\:\mathrm{give} \\ $$$$\mathrm{I}\:=\int\:\frac{\mathrm{2arctan}\left(\mathrm{t}\right)}{\alpha×\frac{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\:+\beta×\frac{\mathrm{2t}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }}×\frac{\mathrm{2dt}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} } \\ $$$$=\mathrm{4}\int\:\:\frac{\mathrm{arctan}\left(\mathrm{t}\right)}{\alpha−\alpha\mathrm{t}^{\mathrm{2}} \:+\mathrm{2}\beta\mathrm{t}}\mathrm{dt}\:=−\mathrm{4}\:\int\:\:\frac{\mathrm{arctan}\left(\mathrm{t}\right)}{\alpha\mathrm{t}^{\mathrm{2}} −\mathrm{2}\beta\mathrm{t}\:−\alpha}\mathrm{dt} \\ $$$$\mathrm{let}\:\mathrm{f}\left(\xi\right)\:=\int\:\:\frac{\mathrm{arctan}\left(\xi\mathrm{t}\right)}{\alpha\mathrm{t}^{\mathrm{2}} −\mathrm{2}\beta\mathrm{t}\:−\alpha}\mathrm{dt}\:\mathrm{we}\:\mathrm{have}\: \\ $$$$\mathrm{f}^{'} \left(\xi\right)\:=\int\:\:\frac{\mathrm{t}}{\left(\mathrm{1}+\xi^{\mathrm{2}} \mathrm{t}^{\mathrm{2}} \right)\left(\alpha\mathrm{t}^{\mathrm{2}} −\mathrm{2}\beta\mathrm{t}\:−\alpha\right)}\mathrm{dt} \\ $$$$=_{\xi\mathrm{t}\:=\mathrm{u}} \:\:\:\int\:\:\frac{\mathrm{u}}{\xi\left(\mathrm{1}+\mathrm{u}^{\mathrm{2}} \right)\left(\alpha\frac{\mathrm{u}^{\mathrm{2}} }{\xi^{\mathrm{2}} }−\mathrm{2}\beta\frac{\mathrm{u}}{\xi}−\alpha\right)}\frac{\mathrm{du}}{\xi} \\ $$$$=\int\:\:\frac{\mathrm{udu}}{\xi^{\mathrm{2}} \left(\mathrm{u}^{\mathrm{2}} +\mathrm{1}\right)\left(\alpha\:\frac{\mathrm{u}^{\mathrm{2}} }{\xi^{\mathrm{2}} }\:−\mathrm{2}\beta\:\frac{\mathrm{u}}{\xi}−\alpha\right)}\:=\int\:\:\frac{\mathrm{udu}}{\left(\mathrm{u}^{\mathrm{2}} \:+\mathrm{1}\right)\left(\alpha\mathrm{u}^{\mathrm{2}} −\mathrm{2}\beta\xi\mathrm{u}\:−\alpha\xi^{\mathrm{2}} \right)} \\ $$$$\mathrm{let}\:\mathrm{decompose}\:\mathrm{F}\left(\mathrm{u}\right)\:=\frac{\mathrm{u}}{\left(\mathrm{u}^{\mathrm{2}} \:+\mathrm{1}\right)\left(\alpha\mathrm{u}^{\mathrm{2}} −\mathrm{2}\beta\xi\mathrm{u}\:−\alpha\xi^{\mathrm{2}} \right)} \\ $$$$\alpha\mathrm{u}^{\mathrm{2}} \:−\mathrm{2}\beta\xi\mathrm{u}\:−\alpha\xi^{\mathrm{2}} \:\rightarrow\Delta^{'} \:=\beta^{\mathrm{2}} \xi^{\mathrm{2}} −\alpha^{\mathrm{2}} \xi^{\mathrm{2}} \:=\xi^{\mathrm{2}} \left(\beta^{\mathrm{2}} −\alpha^{\mathrm{2}} \right)\:=\xi^{\mathrm{2}} \left(\mathrm{b}^{\mathrm{4}} −\mathrm{a}^{\mathrm{4}} \right) \\ $$$$\mathrm{case1}\:\:\mid\mathrm{b}\mid>\mid\mathrm{a}\mid\:\Rightarrow\mathrm{u}_{\mathrm{1}} =\frac{\beta\xi+\xi\sqrt{\beta^{\mathrm{2}} −\alpha^{\mathrm{2}} }}{\alpha}\:\mathrm{and}\:\mathrm{u}_{\mathrm{2}} =\frac{\beta\xi−\xi\sqrt{\beta^{\mathrm{2}} −\alpha^{\mathrm{2}} }}{\alpha}\:\Rightarrow \\ $$$$\mathrm{F}\left(\mathrm{u}\right)\:=\frac{\mathrm{u}}{\left(\mathrm{u}^{\mathrm{2}} \:+\mathrm{1}\right)\alpha\left(\mathrm{u}−\mathrm{u}_{\mathrm{1}} \right)\left(\mathrm{u}−\mathrm{u}_{\mathrm{2}} \right)}\:=\frac{\mathrm{a}_{\mathrm{1}} }{\mathrm{u}−\mathrm{u}_{\mathrm{1}} }\:+\frac{\mathrm{a}_{\mathrm{2}} }{\mathrm{u}−\mathrm{u}_{\mathrm{2}} }\:+\frac{\mathrm{b}_{\mathrm{1}} \mathrm{u}\:+\mathrm{b}_{\mathrm{2}} }{\mathrm{u}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$$\mathrm{eazy}\:\mathrm{to}\:\mathrm{find}\:\mathrm{a}_{\mathrm{i}} \:\mathrm{and}\:\mathrm{b}_{\mathrm{i}} \:\Rightarrow \\ $$$$\mathrm{f}^{'} \left(\xi\right)\:=\mathrm{a}_{\mathrm{1}} \mathrm{ln}\mid\mathrm{u}−\mathrm{u}_{\mathrm{1}} \mid\:+\mathrm{a}_{\mathrm{2}} \mathrm{ln}\mid\mathrm{u}−\mathrm{u}_{\mathrm{2}} \mid\:+\frac{\mathrm{b}_{\mathrm{1}} }{\mathrm{2}}\mathrm{ln}\left(\mathrm{u}^{\mathrm{2}} \:+\mathrm{1}\right)+\mathrm{b}_{\mathrm{2}} \mathrm{arctan}\left(\mathrm{u}\right)\:+\mathrm{c}\:\Rightarrow \\ $$$$\mathrm{f}\left(\xi\right)\:=\mathrm{a}_{\mathrm{1}} \int\:\mathrm{ln}\mid\mathrm{u}−\mathrm{u}_{\mathrm{1}} \mid\mathrm{d}\xi\:+\mathrm{a}_{\mathrm{2}} \int\:\mathrm{ln}\mid\mathrm{u}−\mathrm{u}_{\mathrm{2}} \mid\mathrm{d}\xi\:+\frac{\mathrm{b}_{\mathrm{1}} }{\mathrm{2}}\:\int\:\mathrm{ln}\mid\mathrm{u}^{\mathrm{2}} \:+\mathrm{1}\mid\mathrm{d}\xi\:+\mathrm{b}_{\mathrm{2}} \int\:\mathrm{arctanu}\:\mathrm{d}\xi\:+\mathrm{c}\xi \\ $$$$…\mathrm{be}\:\mathrm{continued}… \\ $$$$ \\ $$

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