Question Number 57666 by maxmathsup by imad last updated on 09/Apr/19
$$\left.\mathrm{1}\right)\:{calculate}\:{f}\left(\theta\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{{t}^{\mathrm{2}} \:+\mathrm{2}{sin}\theta{t}\:+\mathrm{1}}{dt}\:\:\:\:{with}\:\mathrm{0}\leqslant\theta\leqslant\frac{\pi}{\mathrm{2}} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{g}\left({t}\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{{t}^{\mathrm{2}} \:+\mathrm{2}\left({sin}\theta\right){t}\:+\mathrm{1}}{d}\theta \\ $$$$\left.\mathrm{3}\right)\:{find}\:{also}\:{h}\left(\theta\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{{t}}{\sqrt{{t}^{\mathrm{2}} \:+\mathrm{2}\left({sin}\theta\right){t}\:+\mathrm{1}}}{dt} \\ $$
Commented by kaivan.ahmadi last updated on 10/Apr/19
$${thank}\:{sir},\:{it}\:{was}\:{fals} \\ $$
Commented by kaivan.ahmadi last updated on 10/Apr/19
$${but}\:{its}\:{equal}\:{to} \\ $$$$\left({t}+{sin}\theta\right)^{\mathrm{2}} +\mathrm{1}−{sin}^{\mathrm{2}} \theta=\left({t}+{sin}\theta\right)^{\mathrm{2}} +{cos}^{\mathrm{2}} \theta \\ $$
Commented by maxmathsup by imad last updated on 10/Apr/19
![1) we have f(θ) =∫_0 ^1 (√(t^2 +2sinθ t +sin^2 θ +cos^2 ))dt =∫_0 ^1 (√((t+sinθ)^2 +cos^2 θ))dt changement t +sinθ =cosθ u give f(θ) = ∫_(tanθ) ^((1+sinθ)/(cosθ)) cosθ (√(1+u^2 ))cosθ du =cos^2 θ ∫_(tanθ) ^((1+sinθ)/(cosθ)) (√(1+u^2 ))du let find I =∫ (√(1+u^2 ))dy chang.u =shα give I =∫ chαch(α)dα=∫((1+ch(2α))/2)dα =(1/2)α +(1/4)sh(2α) +c =(α/2) +(1/2)sh(α)ch(α) +c =((argsh(u))/2) +(1/2) u(√(1+u^2 )) =(1/2)ln(u+(√(1+u^2 ))) +(u/2)(√(1+u^2 )) +c ⇒ f(θ) =((cos^2 (θ))/2)[ln(u+(√(1+u^2 )))+u(√(1+u^2 ))]_(tanθ) ^((1+sinθ)/(cosθ)) =(1/2) cos^2 θ{ ln(((1+sinθ)/(cosθ)) +(√(1+(((1+cosθ)/(sinθ)))^2 )))+((1+sinθ)/(cosθ))(√(1+(((1+sinθ)/(cosθ)))^2 )) −ln(tanθ +(√(1+tan^2 θ))) −tanθ (√(1+tan^2 θ))} .](Q57711.png)
$$\left.\mathrm{1}\right)\:\:{we}\:{have}\:{f}\left(\theta\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{{t}^{\mathrm{2}} \:+\mathrm{2}{sin}\theta\:{t}\:+{sin}^{\mathrm{2}} \theta\:+{cos}^{\mathrm{2}} }{dt} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\left({t}+{sin}\theta\right)^{\mathrm{2}} \:+{cos}^{\mathrm{2}} \theta}{dt}\:{changement}\:{t}\:+{sin}\theta\:\:={cos}\theta\:{u}\:{give} \\ $$$${f}\left(\theta\right)\:=\:\int_{{tan}\theta} ^{\frac{\mathrm{1}+{sin}\theta}{{cos}\theta}} \:\:{cos}\theta\:\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }{cos}\theta\:{du}\:={cos}^{\mathrm{2}} \theta\:\int_{{tan}\theta} ^{\frac{\mathrm{1}+{sin}\theta}{{cos}\theta}} \:\:\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }{du}\:\:{let}\:{find} \\ $$$${I}\:=\int\:\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }{dy}\:\:\:{chang}.{u}\:={sh}\alpha\:{give}\:{I}\:=\int\:{ch}\alpha{ch}\left(\alpha\right){d}\alpha=\int\frac{\mathrm{1}+{ch}\left(\mathrm{2}\alpha\right)}{\mathrm{2}}{d}\alpha \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\alpha\:+\frac{\mathrm{1}}{\mathrm{4}}{sh}\left(\mathrm{2}\alpha\right)\:+{c}\:=\frac{\alpha}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{2}}{sh}\left(\alpha\right){ch}\left(\alpha\right)\:+{c}\:=\frac{{argsh}\left({u}\right)}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{2}}\:{u}\sqrt{\mathrm{1}+{u}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({u}+\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }\right)\:+\frac{{u}}{\mathrm{2}}\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }\:+{c}\:\Rightarrow \\ $$$${f}\left(\theta\right)\:=\frac{{cos}^{\mathrm{2}} \left(\theta\right)}{\mathrm{2}}\left[{ln}\left({u}+\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }\right)+{u}\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }\right]_{{tan}\theta} ^{\frac{\mathrm{1}+{sin}\theta}{{cos}\theta}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:{cos}^{\mathrm{2}} \theta\left\{\:{ln}\left(\frac{\mathrm{1}+{sin}\theta}{{cos}\theta}\:+\sqrt{\mathrm{1}+\left(\frac{\mathrm{1}+{cos}\theta}{{sin}\theta}\right)^{\mathrm{2}} }\right)+\frac{\mathrm{1}+{sin}\theta}{{cos}\theta}\sqrt{\mathrm{1}+\left(\frac{\mathrm{1}+{sin}\theta}{{cos}\theta}\right)^{\mathrm{2}} }\right. \\ $$$$\left.−{ln}\left({tan}\theta\:+\sqrt{\mathrm{1}+{tan}^{\mathrm{2}} \theta}\right)\:−{tan}\theta\:\sqrt{\mathrm{1}+{tan}^{\mathrm{2}} \theta}\right\}\:. \\ $$