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Question Number 104459 by bemath last updated on 21/Jul/20

∫ (dx/(√(Acos x+B)))

$$\int\:\frac{{dx}}{\sqrt{{A}\mathrm{cos}\:{x}+{B}}} \\ $$

Commented by Dwaipayan Shikari last updated on 21/Jul/20

∫(dx/(√(Acosx+B)))=∫((A(−sinx))/(−Asinx)).(1/(√(Acosx+B)))dx {Acosx+B=t^2 ∫((2tdt)/(t(−Asinx)))=−(2/A)∫(1/(sinx))dt {cosx=(t^2 −B).(1/A) −(2/A)∫ (1/(√(A^2 −(t^2 −B)^2 )))dt {sinx=(√(1−(((t^2 −B)/A))^2 )) −(2/A)∫((2tdt)/(2t(√(A^2 −u^2 )))) {t^2 −B=u −(1/A)∫(du/(√(u+B))).(1/(√(A^2 −u^2 )))....continue

$$\int\frac{{dx}}{\sqrt{{Acosx}+{B}}}=\int\frac{{A}\left(−{sinx}\right)}{−{Asinx}}.\frac{\mathrm{1}}{\sqrt{{Acosx}+{B}}}{dx}\:\:\left\{{Acosx}+{B}={t}^{\mathrm{2}} \right. \\ $$$$\int\frac{\mathrm{2}{tdt}}{{t}\left(−{Asinx}\right)}=−\frac{\mathrm{2}}{{A}}\int\frac{\mathrm{1}}{{sinx}}{dt}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left\{{cosx}=\left({t}^{\mathrm{2}} −{B}\right).\frac{\mathrm{1}}{{A}}\right. \\ $$$$\:−\frac{\mathrm{2}}{{A}}\int\:\frac{\mathrm{1}}{\sqrt{{A}^{\mathrm{2}} −\left({t}^{\mathrm{2}} −{B}\right)^{\mathrm{2}} }}{dt}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left\{{sinx}=\sqrt{\mathrm{1}−\left(\frac{{t}^{\mathrm{2}} −{B}}{{A}}\right)^{\mathrm{2}} }\right. \\ $$$$−\frac{\mathrm{2}}{{A}}\int\frac{\mathrm{2}{tdt}}{\mathrm{2}{t}\sqrt{{A}^{\mathrm{2}} −{u}^{\mathrm{2}} }}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left\{{t}^{\mathrm{2}} −{B}={u}\right. \\ $$$$−\frac{\mathrm{1}}{{A}}\int\frac{{du}}{\sqrt{{u}+{B}}}.\frac{\mathrm{1}}{\sqrt{{A}^{\mathrm{2}} −{u}^{\mathrm{2}} }}....{continue} \\ $$$$ \\ $$$$ \\ $$

Answered by bobhans last updated on 21/Jul/20

I = ∫ (dx/(√(Acos x+B))) . let x = π−t I = ∫ ((−dt)/((√(−Acos t+B)) )) = ∫ ((−dx)/(√(−Acos x+B))) 2I = ∫ (1/(√(Acos x+B))) −(1/(√(−Acos x+B))) dx 2I= ∫ (((√(B−Acos x))−(√(B+Acos x)))/(√(B^2 −A^2 cos^2 x))) dx

$${I}\:=\:\int\:\frac{{dx}}{\sqrt{{A}\mathrm{cos}\:{x}+{B}}}\:.\:{let}\:{x}\:=\:\pi−{t}\: \\ $$$${I}\:=\:\int\:\frac{−{dt}}{\sqrt{−{A}\mathrm{cos}\:{t}+{B}}\:}\:=\:\int\:\frac{−{dx}}{\sqrt{−{A}\mathrm{cos}\:{x}+{B}}} \\ $$$$\mathrm{2}{I}\:=\:\int\:\frac{\mathrm{1}}{\sqrt{{A}\mathrm{cos}\:{x}+{B}}}\:−\frac{\mathrm{1}}{\sqrt{−{A}\mathrm{cos}\:{x}+{B}}}\:{dx} \\ $$$$\mathrm{2}{I}=\:\int\:\frac{\sqrt{{B}−{A}\mathrm{cos}\:{x}}−\sqrt{{B}+{A}\mathrm{cos}\:{x}}}{\sqrt{{B}^{\mathrm{2}} −{A}^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} {x}}}\:{dx} \\ $$$$ \\ $$

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