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Question Number 83042 by ~blr237~ last updated on 27/Feb/20
let a,b two positive reals such as a^2 −b^2 =ab Explicit f(a,b)=∫_0 ^(π/2) (du/( (√(a+bsin^2 u))))
$${let}\:\:{a},{b}\:{two}\:{positive}\:{reals}\:{such}\:{as}\:\:{a}^{\mathrm{2}} −{b}^{\mathrm{2}} ={ab} \\ $$$${Explicit}\:\:\:{f}\left({a},{b}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{{du}}{\:\sqrt{{a}+{bsin}^{\mathrm{2}} {u}}}\: \\ $$
Commented by mathmax by abdo last updated on 27/Feb/20
let take a try this integral is eliptic let ϕ(a)=∫_0 ^(π/2) (√(a+bsin^2 x))dx we have ϕ^′ (a) =∫_0 ^(π/2) (dx/(2(√(a+bsin^2 x)))) =((f(a,b))/2) ⇒f(a,b) =2ϕ(a) ϕ(a) =∫_0 ^(π/2) (√(a+b(1−cos^2 x)))dx=∫_0 ^(π/2) (√(a+b−bcos^2 x))dx =∫_0 ^(π/2) (√(a+b−b×(1/(1+tan^2 x))))dx =∫_0 ^(π/2) (√(a+b+(a+b)tan^2 x−b))×(dx/( (√(1+tan^2 x)))) =∫_0 ^(π/2) (√(a+((ab)/(a−b))tan^2 x))×(dx/( (√(1+tan^2 x)))) =∫_0 ^(π/2) (√(a^2 −ab +abtan^2 x))×(dx/( (√(1+tan^2 x)))) =∫_0 ^(π/2) (√(b^2 +abtan^2 x))×(dx/( (√(1+tan^2 x)))) integral at form A_λ =∫_0 ^(π/2) (√((1+λtan^2 x)/(1+tan^2 x)))dx changement tanx =t give A_λ =∫_0 ^∞ (√((1+λt^2 )/(1+t^2 )))(dt/(1+t^2 )) also chang.(√((1+λt^2 )/(1+t^2 )))=u give ((1+λt^2 )/(1+t^2 ))=u^2 ⇒1+λt^2 =u^2 +u^2 t^2 ⇒(λ−u^2 )t^2 =u^2 −1 ⇒ t^2 =((u^2 −1)/(λ−u^2 )) ....be continued....
$${let}\:{take}\:{a}\:{try}\:\:{this}\:{integral}\:{is}\:{eliptic}\:\:{let} \\ $$$$\varphi\left({a}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \sqrt{{a}+{bsin}^{\mathrm{2}} {x}}{dx}\:\:{we}\:{have}\:\varphi^{'} \left({a}\right)\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{dx}}{\mathrm{2}\sqrt{{a}+{bsin}^{\mathrm{2}} {x}}} \\ $$$$=\frac{{f}\left({a},{b}\right)}{\mathrm{2}}\:\Rightarrow{f}\left({a},{b}\right)\:=\mathrm{2}\varphi\left({a}\right) \\ $$$$\varphi\left({a}\right)\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \sqrt{{a}+{b}\left(\mathrm{1}−{cos}^{\mathrm{2}} {x}\right)}{dx}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \sqrt{{a}+{b}−{bcos}^{\mathrm{2}} {x}}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \sqrt{{a}+{b}−{b}×\frac{\mathrm{1}}{\mathrm{1}+{tan}^{\mathrm{2}} {x}}}{dx}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \sqrt{{a}+{b}+\left({a}+{b}\right){tan}^{\mathrm{2}} {x}−{b}}×\frac{{dx}}{\:\sqrt{\mathrm{1}+{tan}^{\mathrm{2}} {x}}} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \sqrt{{a}+\frac{{ab}}{{a}−{b}}{tan}^{\mathrm{2}} {x}}×\frac{{dx}}{\:\sqrt{\mathrm{1}+{tan}^{\mathrm{2}} {x}}} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \sqrt{{a}^{\mathrm{2}} −{ab}\:+{abtan}^{\mathrm{2}} {x}}×\frac{{dx}}{\:\sqrt{\mathrm{1}+{tan}^{\mathrm{2}} {x}}} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \sqrt{{b}^{\mathrm{2}} \:+{abtan}^{\mathrm{2}} {x}}×\frac{{dx}}{\:\sqrt{\mathrm{1}+{tan}^{\mathrm{2}} {x}}}\:\:{integral}\:{at}\:{form} \\ $$$${A}_{\lambda} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \sqrt{\frac{\mathrm{1}+\lambda{tan}^{\mathrm{2}} {x}}{\mathrm{1}+{tan}^{\mathrm{2}} {x}}}{dx}\:\:{changement}\:{tanx}\:={t}\:{give} \\ $$$${A}_{\lambda} =\int_{\mathrm{0}} ^{\infty} \:\sqrt{\frac{\mathrm{1}+\lambda{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }\:\:{also}\:{chang}.\sqrt{\frac{\mathrm{1}+\lambda{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}={u}\:{give} \\ $$$$\frac{\mathrm{1}+\lambda{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }={u}^{\mathrm{2}} \:\Rightarrow\mathrm{1}+\lambda{t}^{\mathrm{2}} ={u}^{\mathrm{2}} \:+{u}^{\mathrm{2}} \:{t}^{\mathrm{2}} \:\Rightarrow\left(\lambda−{u}^{\mathrm{2}} \right){t}^{\mathrm{2}} ={u}^{\mathrm{2}} −\mathrm{1}\:\Rightarrow \\ $$$${t}^{\mathrm{2}} =\frac{{u}^{\mathrm{2}} −\mathrm{1}}{\lambda−{u}^{\mathrm{2}} }\:….{be}\:{continued}…. \\ $$$$ \\ $$

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