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Question Number 130088 by stelor last updated on 22/Jan/21
hello please. ∫(dx/(a^2 cos^2 x + b^2 sin^2 x))
$$\mathrm{hello}\:\mathrm{please}. \\ $$$$\:\:\int\frac{{dx}}{{a}^{\mathrm{2}} {cos}^{\mathrm{2}} {x}\:+\:{b}^{\mathrm{2}} {sin}^{\mathrm{2}} {x}} \\ $$
Answered by bobhans last updated on 22/Jan/21
I=∫ ((dx/(cos^2 x))/(a^2 +b^2 tan^2 x)) = ∫ ((sec^2 x dx)/(a^2 +b^2 tan^2 x)) let tan x = u ⇒du = sec^2 x dx I=∫ (du/(a^2 +b^2 u^2 )) ; let bu = a tan θ I=∫ (((a/b)sec^2 θ dθ)/(a^2 sec^2 θ)) = (1/(ab)) θ + c I=(1/(ab)) tan^(−1) (((bu)/a))+c = (1/(ab))tan^(−1) (((b tan x)/a))+c
$${I}=\int\:\frac{\frac{{dx}}{\mathrm{cos}\:^{\mathrm{2}} {x}}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} \mathrm{tan}\:^{\mathrm{2}} {x}}\:=\:\int\:\frac{\mathrm{sec}\:^{\mathrm{2}} {x}\:{dx}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} \:\mathrm{tan}\:^{\mathrm{2}} {x}} \\ $$$$\:{let}\:\mathrm{tan}\:{x}\:=\:{u}\:\Rightarrow{du}\:=\:\mathrm{sec}\:^{\mathrm{2}} {x}\:{dx} \\ $$$${I}=\int\:\frac{{du}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} {u}^{\mathrm{2}} }\:\:;\:{let}\:{bu}\:=\:{a}\:\mathrm{tan}\:\theta \\ $$$${I}=\int\:\frac{\frac{{a}}{{b}}\mathrm{sec}\:^{\mathrm{2}} \theta\:{d}\theta}{{a}^{\mathrm{2}} \mathrm{sec}\:^{\mathrm{2}} \theta}\:=\:\frac{\mathrm{1}}{{ab}}\:\theta\:+\:{c} \\ $$$${I}=\frac{\mathrm{1}}{{ab}}\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{{bu}}{{a}}\right)+{c}\:=\:\frac{\mathrm{1}}{{ab}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{{b}\:\mathrm{tan}\:{x}}{{a}}\right)+{c}\: \\ $$
Commented by stelor last updated on 22/Jan/21
thank...
$$\mathrm{thank}… \\ $$
Answered by mathmax by abdo last updated on 22/Jan/21
I =∫ (dx/(a^2 (cos^2 x+(b^2 /a^2 )sin^2 x))) let α =(b^2 /a^2 ) (ab≠0and a≠b)) ⇒I=(1/a^2 )∫(dx/(cos^2 x+α^2 cos^2 x)) =(1/a^2 )∫ (dx/(((1+cos(2x))/2)+α^2 ×((1−cos(2x))/2))) =(2/a^2 )∫ (dx/(1+cos(2x)+α^2 −α^2 cos(2x))) =(2/a^2 )∫ (dx/(1+α^2 +(1−α^2 )cos(2x))) =_(2x=t) (1/a^2 )∫ (dt/((1+α^2 +(1−α^2 )cost))) =_(tan((t/2))=z) (1/a^2 )∫ ((2dz)/((1+z^2 ){1+α^2 +(1−α^2 )((1−z^2 )/(1+z^2 ))})) =(2/a^2 )∫ (dz/((1+α^2 )(1+z^2 )+(1−α^2 )(1−z^2 ))) =(2/a^2 )∫ (dz/(1+α^2 +(1+α^2 )z^2 +1−α^2 −(1−α^2 )z^2 )) =(2/a^2 )∫ (dz/(2+(2α^2 )z^2 )) =(1/a^2 )∫ (dz/(1+α^2 z^2 )) =(1/(αa^2 )) arctan(αz) +c =(1/(αa^2 ))arctan(αtan((t/2)))+C =(a^2 /(b^2 a^2 ))arctan((b^2 /a^2 )tan(x))+C =(1/b^2 )arctan((b^2 /a^2 )tan(x)) +C
$$\left.\mathrm{I}\:=\int\:\:\frac{\mathrm{dx}}{\mathrm{a}^{\mathrm{2}} \left(\mathrm{cos}^{\mathrm{2}} \mathrm{x}+\frac{\mathrm{b}^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}} }\mathrm{sin}^{\mathrm{2}} \mathrm{x}\right)}\:\mathrm{let}\:\alpha\:=\frac{\mathrm{b}^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}} }\:\:\left(\mathrm{ab}\neq\mathrm{0and}\:\mathrm{a}\neq\mathrm{b}\right)\right)\:\Rightarrow\mathrm{I}=\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{2}} }\int\frac{\mathrm{dx}}{\mathrm{cos}^{\mathrm{2}} \mathrm{x}+\alpha^{\mathrm{2}} \:\mathrm{cos}^{\mathrm{2}} \mathrm{x}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{2}} }\int\:\:\frac{\mathrm{dx}}{\frac{\mathrm{1}+\mathrm{cos}\left(\mathrm{2x}\right)}{\mathrm{2}}+\alpha^{\mathrm{2}} ×\frac{\mathrm{1}−\mathrm{cos}\left(\mathrm{2x}\right)}{\mathrm{2}}}\:=\frac{\mathrm{2}}{\mathrm{a}^{\mathrm{2}} }\int\:\frac{\mathrm{dx}}{\mathrm{1}+\mathrm{cos}\left(\mathrm{2x}\right)+\alpha^{\mathrm{2}} −\alpha^{\mathrm{2}} \mathrm{cos}\left(\mathrm{2x}\right)} \\ $$$$=\frac{\mathrm{2}}{\mathrm{a}^{\mathrm{2}} }\int\:\frac{\mathrm{dx}}{\mathrm{1}+\alpha^{\mathrm{2}} \:+\left(\mathrm{1}−\alpha^{\mathrm{2}} \:\right)\mathrm{cos}\left(\mathrm{2x}\right)}\:=_{\mathrm{2x}=\mathrm{t}} \:\:\:\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{2}} }\int\:\frac{\mathrm{dt}}{\left(\mathrm{1}+\alpha^{\mathrm{2}} \:+\left(\mathrm{1}−\alpha^{\mathrm{2}} \right)\mathrm{cost}\right)} \\ $$$$=_{\mathrm{tan}\left(\frac{\mathrm{t}}{\mathrm{2}}\right)=\mathrm{z}} \:\:\:\:\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{2}} }\int\:\:\frac{\mathrm{2dz}}{\left(\mathrm{1}+\mathrm{z}^{\mathrm{2}} \right)\left\{\mathrm{1}+\alpha^{\mathrm{2}} \:+\left(\mathrm{1}−\alpha^{\mathrm{2}} \right)\frac{\mathrm{1}−\mathrm{z}^{\mathrm{2}} }{\mathrm{1}+\mathrm{z}^{\mathrm{2}} }\right\}} \\ $$$$=\frac{\mathrm{2}}{\mathrm{a}^{\mathrm{2}} }\int\:\:\:\frac{\mathrm{dz}}{\left(\mathrm{1}+\alpha^{\mathrm{2}} \right)\left(\mathrm{1}+\mathrm{z}^{\mathrm{2}} \right)+\left(\mathrm{1}−\alpha^{\mathrm{2}} \right)\left(\mathrm{1}−\mathrm{z}^{\mathrm{2}} \right)} \\ $$$$=\frac{\mathrm{2}}{\mathrm{a}^{\mathrm{2}} }\int\:\frac{\mathrm{dz}}{\mathrm{1}+\alpha^{\mathrm{2}} \:+\left(\mathrm{1}+\alpha^{\mathrm{2}} \right)\mathrm{z}^{\mathrm{2}} \:+\mathrm{1}−\alpha^{\mathrm{2}} −\left(\mathrm{1}−\alpha^{\mathrm{2}} \right)\mathrm{z}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{2}}{\mathrm{a}^{\mathrm{2}} }\int\:\frac{\mathrm{dz}}{\mathrm{2}+\left(\mathrm{2}\alpha^{\mathrm{2}} \right)\mathrm{z}^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{2}} }\int\:\:\frac{\mathrm{dz}}{\mathrm{1}+\alpha^{\mathrm{2}} \mathrm{z}^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\alpha\mathrm{a}^{\mathrm{2}} }\:\mathrm{arctan}\left(\alpha\mathrm{z}\right)\:+\mathrm{c} \\ $$$$=\frac{\mathrm{1}}{\alpha\mathrm{a}^{\mathrm{2}} }\mathrm{arctan}\left(\alpha\mathrm{tan}\left(\frac{\mathrm{t}}{\mathrm{2}}\right)\right)+\mathrm{C} \\ $$$$=\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{b}^{\mathrm{2}} \mathrm{a}^{\mathrm{2}} }\mathrm{arctan}\left(\frac{\mathrm{b}^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}} }\mathrm{tan}\left(\mathrm{x}\right)\right)+\mathrm{C} \\ $$$$=\frac{\mathrm{1}}{\mathrm{b}^{\mathrm{2}} }\mathrm{arctan}\left(\frac{\mathrm{b}^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}} }\mathrm{tan}\left(\mathrm{x}\right)\right)\:+\mathrm{C} \\ $$
Commented by mathmax by abdo last updated on 22/Jan/21
sorry α=(b/a) ⇒I =(a/(ba^2 )) arctan((b/a)tan(x)) +C =(1/(ab))arctan((b/a)tanx)+C
$$\mathrm{sorry}\:\alpha=\frac{\mathrm{b}}{\mathrm{a}}\:\Rightarrow\mathrm{I}\:=\frac{\mathrm{a}}{\mathrm{ba}^{\mathrm{2}} }\:\mathrm{arctan}\left(\frac{\mathrm{b}}{\mathrm{a}}\mathrm{tan}\left(\mathrm{x}\right)\right)\:+\mathrm{C} \\ $$$$=\frac{\mathrm{1}}{\mathrm{ab}}\mathrm{arctan}\left(\frac{\mathrm{b}}{\mathrm{a}}\mathrm{tanx}\right)+\mathrm{C} \\ $$

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