Question Number 53463 by maxmathsup by imad last updated on 22/Jan/19
$$\left.\mathrm{1}\right){let}\:\mathrm{0}<\theta<\frac{\pi}{\mathrm{2}}\:\:\:\:{and}\:\:{A}\left(\theta\right)\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{dx}}{\:\sqrt{{x}^{\mathrm{2}} \:+\mathrm{2}{sin}\theta\:{x}\:+\mathrm{1}}} \\ $$$${calculate}\:{A}\left(\theta\right) \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{dx}}{\:\sqrt{{x}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}{x}\:+\mathrm{1}}} \\ $$
Commented by maxmathsup by imad last updated on 23/Jan/19
![1) we have A(θ) =∫_0 ^(π/2) (dx/( (√(x^2 +2x sinθ +sin^2 θ +cos^2 θ)))) =∫_0 ^(π/2) (dx/( (√((x+sinθ)^(2 ) +cos^2 θ)))) changement x+sinθ=t cosθ give t =((x+sinθ)/(cosθ)) A(θ) = ∫_(tanθ) ^((π/(2cosθ))+tanθ) ((cosθ dt)/(cosθ(√(1+t^2 )))) = ∫_(tanθ) ^((π/(2cosθ)) +tanθ) (dt/( (√(1+t^2 )))) =[argsh(t)]_(tanθ) ^((π/(2cosθ))+tanθ) =ln(t+(√(1+t^2 ))]_(tanθ) ^((π/(2cosθ)) +tanθ) = ln((π/(2cosθ)) +tanθ +(√(1+((π/(2cosθ))+tanθ)^2 ))) −ln(tanθ +(√(1+tan^2 θ))) 2) ∫_0 ^(π/2) (dx/( (√(x^2 +(√2)x +1)))) =A((π/4)) =ln((π/(2 cos((π/4)))) +tan((π/4))+(√(1+((π/(2cos((π/4))))+tan((π/4))^2 )) −ln(tan((π/4))+(√(1+tan^2 ((π/4))))) =ln((π/( (√2))) +1 +(√(1+((π/( (√2)))+1)^2 )))−ln(1+(√2)).](https://www.tinkutara.com/question/Q53603.png)
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{A}\left(\theta\right)\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\frac{{dx}}{\:\sqrt{{x}^{\mathrm{2}} \:+\mathrm{2}{x}\:{sin}\theta\:+{sin}^{\mathrm{2}} \theta\:+{cos}^{\mathrm{2}} \theta}} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{dx}}{\:\sqrt{\left({x}+{sin}\theta\right)^{\mathrm{2}\:} +{cos}^{\mathrm{2}} \theta}}\:\:{changement}\:{x}+{sin}\theta={t}\:{cos}\theta\:{give}\:{t}\:=\frac{{x}+{sin}\theta}{{cos}\theta} \\ $$$${A}\left(\theta\right)\:=\:\int_{{tan}\theta} ^{\frac{\pi}{\mathrm{2}{cos}\theta}+{tan}\theta} \:\:\:\:\:\frac{{cos}\theta\:{dt}}{{cos}\theta\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}\:=\:\int_{{tan}\theta} ^{\frac{\pi}{\mathrm{2}{cos}\theta}\:+{tan}\theta} \:\:\frac{{dt}}{\:\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}\:=\left[{argsh}\left({t}\right)\right]_{{tan}\theta} ^{\frac{\pi}{\mathrm{2}{cos}\theta}+{tan}\theta} \\ $$$$={ln}\left({t}+\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\right]_{{tan}\theta} ^{\frac{\pi}{\mathrm{2}{cos}\theta}\:+{tan}\theta} \:=\:{ln}\left(\frac{\pi}{\mathrm{2}{cos}\theta}\:+{tan}\theta\:+\sqrt{\mathrm{1}+\left(\frac{\pi}{\mathrm{2}{cos}\theta}+{tan}\theta\right)^{\mathrm{2}} }\right) \\ $$$$−{ln}\left({tan}\theta\:+\sqrt{\mathrm{1}+{tan}^{\mathrm{2}} \theta}\right) \\ $$$$\left.\mathrm{2}\right)\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{dx}}{\:\sqrt{{x}^{\mathrm{2}} +\sqrt{\mathrm{2}}{x}\:+\mathrm{1}}}\:={A}\left(\frac{\pi}{\mathrm{4}}\right)\:={ln}\left(\frac{\pi}{\mathrm{2}\:{cos}\left(\frac{\pi}{\mathrm{4}}\right)}\:+{tan}\left(\frac{\pi}{\mathrm{4}}\right)+\sqrt{\mathrm{1}+\left(\frac{\pi}{\mathrm{2}{cos}\left(\frac{\pi}{\mathrm{4}}\right)}+{tan}\left(\frac{\pi}{\mathrm{4}}\right)^{\mathrm{2}} \right.}\right. \\ $$$$−{ln}\left({tan}\left(\frac{\pi}{\mathrm{4}}\right)+\sqrt{\mathrm{1}+{tan}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{4}}\right)}\right) \\ $$$$={ln}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\:+\mathrm{1}\:+\sqrt{\mathrm{1}+\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}+\mathrm{1}\right)^{\mathrm{2}} }\right)−{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right). \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 22/Jan/19
$$\left.\mathrm{1}\right)\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{dx}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{xsin}\theta+\mathrm{1}}} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{dx}}{\:\sqrt{\left({x}+{sin}\theta\right)^{\mathrm{2}} +{cos}^{\mathrm{2}} \theta}}=\mid{ln}\left\{\mid\left({x}+{sin}\theta\right)+\sqrt{\left({x}+{sin}\theta\right)^{\mathrm{2}} +{cos}^{\mathrm{2}} \theta}\:\right)\mid_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \\ $$$$=\left[{ln}\left\{\left(\frac{\pi}{\mathrm{2}}+{sin}\theta\right)+\sqrt{\left(\frac{\pi}{\mathrm{2}}+{sin}\theta\right)^{\mathrm{2}} +{cos}^{\mathrm{2}} \theta}\:\right\}−{ln}\left\{\left(\mathrm{0}+{sin}\theta\right)+\sqrt{\left(\mathrm{0}+{sin}\theta\right)^{\mathrm{2}} +{cos}^{\mathrm{2}} \theta}\:\right\}\right. \\ $$$$={ln}\left\{\left(\frac{\pi}{\mathrm{2}}+{sin}\theta\right)+\sqrt{\frac{\pi^{\mathrm{2}} }{\mathrm{4}}+\pi{sin}\theta+\mathrm{1}}\:\right\}−{ln}\left\{{sin}\theta+\mathrm{1}\right\} \\ $$$$\:{using}\:{formula} \\ $$$$\int\frac{{dx}}{\:\sqrt{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }}={ln}\left({x}+\sqrt{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }\:\right) \\ $$$$\left.\mathrm{2}\right){put}\:\theta=\frac{\pi}{\mathrm{4}}\:{for}\:{answer}… \\ $$$${ln}\left\{\left(\frac{\pi}{\mathrm{4}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)+\sqrt{\frac{\pi^{\mathrm{2}} }{\mathrm{4}}+\frac{\pi}{\:\sqrt{\mathrm{2}}}+\mathrm{1}}\:\right\}−{ln}\left\{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\mathrm{1}\right\} \\ $$$${sir}\:{pls}\:{check}… \\ $$