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Question Number 40823 by math khazana by abdo last updated on 28/Jul/18
calculate ∫_0 ^(π/2) (√(cos^2 x +3sin^2 x))dx
$${calculate}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \sqrt{{cos}^{\mathrm{2}} {x}\:+\mathrm{3}{sin}^{\mathrm{2}} {x}}{dx} \\ $$
Commented by maxmathsup by imad last updated on 31/Jul/18
let I = ∫_0 ^(π/2) (√(cos^2 x +3sin^2 x))dx I = ∫_0 ^(π/2) cosx(√(1+3tan^2 x))dx= ∫_0 ^(π/2) ((√(1+3tan^2 x))/( (√(1+tan^2 x))))dx ch (√3)tanx =sh(u) =∫_0 ^∞ ((√(1+sh^2 u))/( (√(1+((sh^2 u)/3)))))(((1/( (√3)))chu)/((1+((sh^2 u)/3)))) du (x=arctan(((shu)/( (√3))))) =(1/( (√3))) ∫_0 ^∞ ((ch^2 u du)/( (√(3+sh^2 u))(3+sh^2 u))) 3(√3)du= 3 ∫_0 ^∞ ((ch^2 u)/((3+sh^2 u)^(3/2) )) du = 3 ∫_0 ^∞ (((1+ch(2u))/2)/((3+((ch(2u)−1)/2))^(3/2) )) du = (3/2) 2^(3/2) ∫_0 ^∞ ((1+ch(2u))/((5+ch(2u))^(3/2) )) du =3(√2) ∫_0 ^∞ ((1+((e^(2u) +e^(−2u) )/2))/((5 +((e^(2u) +e^(−2u) )/2))^(3/2) )) du =3(√2).(2^(3/2) /2) ∫_0 ^∞ ((2 +e^(2u) +e^(−2u) )/((10 +e^(2u) +e^(−2u) )^(3/2) )) du =_(e^u =t) 6 ∫_1 ^(+∞) ((2 +t^2 +t^(−2) )/((10 +t^2 +t^(−2) )^(3/2) )) (dt/t) =6 ∫_1 ^(+∞) ((2t^2 +t^4 +1)/(t^3 (((10t^2 +t^4 +1)/t^2 ))^(3/2) ))dt = 6 ∫_0 ^∞ ((2t^2 +t^4 +1)/( (t^4 +10t^2 +1)^(3/2) )) dt ...be continued...
$${let}\:{I}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \sqrt{{cos}^{\mathrm{2}} {x}\:+\mathrm{3}{sin}^{\mathrm{2}} {x}}{dx} \\ $$$${I}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {cosx}\sqrt{\mathrm{1}+\mathrm{3}{tan}^{\mathrm{2}} {x}}{dx}=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{\sqrt{\mathrm{1}+\mathrm{3}{tan}^{\mathrm{2}} {x}}}{\:\sqrt{\mathrm{1}+{tan}^{\mathrm{2}} {x}}}{dx}\:\:{ch}\:\:\:\:\:\sqrt{\mathrm{3}}{tanx}\:={sh}\left({u}\right) \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{\sqrt{\mathrm{1}+{sh}^{\mathrm{2}} {u}}}{\:\sqrt{\mathrm{1}+\frac{{sh}^{\mathrm{2}} {u}}{\mathrm{3}}}}\frac{\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}{chu}}{\left(\mathrm{1}+\frac{{sh}^{\mathrm{2}} {u}}{\mathrm{3}}\right)}\:{du}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({x}={arctan}\left(\frac{{shu}}{\:\sqrt{\mathrm{3}}}\right)\right) \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{ch}^{\mathrm{2}} {u}\:{du}}{\:\sqrt{\mathrm{3}+{sh}^{\mathrm{2}} {u}}\left(\mathrm{3}+{sh}^{\mathrm{2}} {u}\right)}\:\mathrm{3}\sqrt{\mathrm{3}}{du}=\:\mathrm{3}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{ch}^{\mathrm{2}} {u}}{\left(\mathrm{3}+{sh}^{\mathrm{2}} {u}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }\:{du} \\ $$$$=\:\mathrm{3}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\frac{\mathrm{1}+{ch}\left(\mathrm{2}{u}\right)}{\mathrm{2}}}{\left(\mathrm{3}+\frac{{ch}\left(\mathrm{2}{u}\right)−\mathrm{1}}{\mathrm{2}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }\:{du}\:=\:\frac{\mathrm{3}}{\mathrm{2}}\:\mathrm{2}^{\frac{\mathrm{3}}{\mathrm{2}}} \:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{1}+{ch}\left(\mathrm{2}{u}\right)}{\left(\mathrm{5}+{ch}\left(\mathrm{2}{u}\right)\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }\:{du} \\ $$$$=\mathrm{3}\sqrt{\mathrm{2}}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{\mathrm{1}+\frac{{e}^{\mathrm{2}{u}} \:+{e}^{−\mathrm{2}{u}} }{\mathrm{2}}}{\left(\mathrm{5}\:+\frac{{e}^{\mathrm{2}{u}} \:+{e}^{−\mathrm{2}{u}} }{\mathrm{2}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }\:{du} \\ $$$$=\mathrm{3}\sqrt{\mathrm{2}}.\frac{\mathrm{2}^{\frac{\mathrm{3}}{\mathrm{2}}} }{\mathrm{2}}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\:\frac{\mathrm{2}\:+{e}^{\mathrm{2}{u}} \:+{e}^{−\mathrm{2}{u}} }{\left(\mathrm{10}\:+{e}^{\mathrm{2}{u}} \:+{e}^{−\mathrm{2}{u}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }\:{du} \\ $$$$=_{{e}^{{u}} ={t}} \:\:\mathrm{6}\:\:\int_{\mathrm{1}} ^{+\infty} \:\:\:\:\frac{\mathrm{2}\:+{t}^{\mathrm{2}} \:+{t}^{−\mathrm{2}} }{\left(\mathrm{10}\:+{t}^{\mathrm{2}} \:+{t}^{−\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }\:\frac{{dt}}{{t}} \\ $$$$=\mathrm{6}\:\int_{\mathrm{1}} ^{+\infty} \:\:\:\:\:\frac{\mathrm{2}{t}^{\mathrm{2}} \:+{t}^{\mathrm{4}} \:+\mathrm{1}}{{t}^{\mathrm{3}} \left(\frac{\mathrm{10}{t}^{\mathrm{2}} \:+{t}^{\mathrm{4}} \:+\mathrm{1}}{{t}^{\mathrm{2}} }\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{dt}\:=\:\mathrm{6}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{\mathrm{2}{t}^{\mathrm{2}} \:+{t}^{\mathrm{4}} \:+\mathrm{1}}{\:\left({t}^{\mathrm{4}} \:+\mathrm{10}{t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }\:{dt}\:\:…{be}\:{continued}… \\ $$$$ \\ $$

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