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Question Number 60893 by mathsolverby Abdo last updated on 26/May/19

calculate ∫_0 ^(π/2) (dx/((√2)cos^2 x +(√3)sin^2 x))

$${calculate}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{dx}}{\sqrt{\mathrm{2}}{cos}^{\mathrm{2}} {x}\:+\sqrt{\mathrm{3}}{sin}^{\mathrm{2}} {x}} \\ $$

Commented by maxmathsup by imad last updated on 27/May/19

let A =∫_0 ^(π/2) (dx/((√2)cos^2 x +(√3)sin^2 x)) ⇒ A =∫_0 ^(π/2) (dx/((√2)((1+cos(2x))/2) +(√3)((1−cos(2x))/2))) =2 ∫_0 ^(π/2) (dx/((√2) +(√3) +((√2)−(√3))cos(2x))) =_(tanx=t) 2 ∫_0 ^(+∞) (1/((√2) +(√3)+((√2)−(√3))((1−t^2 )/(1+t^2 )))) (dt/(1+t^2 )) =2 ∫_0 ^∞ (1/((1+t^2 )((√2) +(√3)) +(√2)−(√3))) dt =2 ∫_0 ^∞ (dt/((√2) +(√3) +((√2) +(√3))t^2 +(√2)−(√3))) =2∫_0 ^∞ (dt/(2(√2) +((√2)+(√3))t^2 )) =(1/(√2))∫_0 ^∞ (dt/(1+(((√2) +(√3))/(2(√2)))t^2 )) changement (√(((√2) +(√3))/(2(√2) )))t =u give A =(1/(√2)) ∫_0 ^∞ (1/(1+u^2 )) (√((2(√2))/((√2) +(√3)))) du =((√(√2))/(√((√2) +(√3)))) ∫_0 ^∞ (du/(1+u^2 )) =(π/2) ((√(√2))/(√((√2) +(√3)))) .

$${let}\:{A}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{dx}}{\sqrt{\mathrm{2}}{cos}^{\mathrm{2}} {x}\:+\sqrt{\mathrm{3}}{sin}^{\mathrm{2}} {x}}\:\Rightarrow\:{A}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{dx}}{\sqrt{\mathrm{2}}\frac{\mathrm{1}+{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}}\:+\sqrt{\mathrm{3}}\frac{\mathrm{1}−{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}}} \\ $$$$=\mathrm{2}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{dx}}{\sqrt{\mathrm{2}}\:+\sqrt{\mathrm{3}}\:+\left(\sqrt{\mathrm{2}}−\sqrt{\mathrm{3}}\right){cos}\left(\mathrm{2}{x}\right)}\:=_{{tanx}={t}} \:\:\:\mathrm{2}\:\int_{\mathrm{0}} ^{+\infty} \:\:\:\frac{\mathrm{1}}{\sqrt{\mathrm{2}}\:+\sqrt{\mathrm{3}}+\left(\sqrt{\mathrm{2}}−\sqrt{\mathrm{3}}\right)\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}\:\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$=\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{\mathrm{1}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left(\sqrt{\mathrm{2}}\:+\sqrt{\mathrm{3}}\right)\:+\sqrt{\mathrm{2}}−\sqrt{\mathrm{3}}}\:{dt} \\ $$$$=\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\:\frac{{dt}}{\sqrt{\mathrm{2}}\:+\sqrt{\mathrm{3}}\:+\left(\sqrt{\mathrm{2}}\:+\sqrt{\mathrm{3}}\right){t}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}−\sqrt{\mathrm{3}}}\:=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{{dt}}{\mathrm{2}\sqrt{\mathrm{2}}\:+\left(\sqrt{\mathrm{2}}+\sqrt{\mathrm{3}}\right){t}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{dt}}{\mathrm{1}+\frac{\sqrt{\mathrm{2}}\:+\sqrt{\mathrm{3}}}{\mathrm{2}\sqrt{\mathrm{2}}}{t}^{\mathrm{2}} }\:\:\:{changement}\:\:\sqrt{\frac{\sqrt{\mathrm{2}}\:+\sqrt{\mathrm{3}}}{\mathrm{2}\sqrt{\mathrm{2}}\:}}{t}\:={u}\:{give} \\ $$$${A}\:=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{1}+{u}^{\mathrm{2}} }\:\sqrt{\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\sqrt{\mathrm{2}}\:+\sqrt{\mathrm{3}}}}\:{du}\:=\frac{\sqrt{\sqrt{\mathrm{2}}}}{\sqrt{\sqrt{\mathrm{2}}\:+\sqrt{\mathrm{3}}}}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{du}}{\mathrm{1}+{u}^{\mathrm{2}} }\:=\frac{\pi}{\mathrm{2}}\:\frac{\sqrt{\sqrt{\mathrm{2}}}}{\sqrt{\sqrt{\mathrm{2}}\:+\sqrt{\mathrm{3}}}}\:. \\ $$

Answered by Prithwish sen last updated on 27/May/19

∫(dx/((√2)cos^2 x+ (√3)sin^2 x)) =∫((sec^2 x dx)/((√2)+(√(3 ))tan^2 x)) Now putting tanx = v then sec^2 x dx =dv the integral changes to ∫(dv/((√2)+(√3)v^2 ))

$$\int\frac{\mathrm{dx}}{\sqrt{\mathrm{2}}\mathrm{cos}^{\mathrm{2}} \mathrm{x}+\:\sqrt{\mathrm{3}}\mathrm{sin}^{\mathrm{2}} \mathrm{x}}\:=\int\frac{\mathrm{sec}^{\mathrm{2}} \mathrm{x}\:\mathrm{dx}}{\sqrt{\mathrm{2}}+\sqrt{\mathrm{3}\:}\mathrm{tan}^{\mathrm{2}} \mathrm{x}} \\ $$$$\mathrm{Now}\:\mathrm{putting}\:\mathrm{tanx}\:=\:\mathrm{v} \\ $$$$\mathrm{then}\:\mathrm{sec}^{\mathrm{2}} \mathrm{x}\:\mathrm{dx}\:=\mathrm{dv} \\ $$$$\:\mathrm{the}\:\mathrm{integral}\:\mathrm{changes}\:\mathrm{to} \\ $$$$\int\frac{\mathrm{dv}}{\sqrt{\mathrm{2}}+\sqrt{\mathrm{3}}\mathrm{v}^{\mathrm{2}} }\: \\ $$

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