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