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Question Number 53463 by maxmathsup by imad last updated on 22/Jan/19

1)let 0<θ<(π/2)    and  A(θ) =∫_0 ^(π/2)   (dx/(√(x^2  +2sinθ x +1)))  calculate A(θ)  2) calculate ∫_0 ^(π/2)   (dx/(√(x^2  +(√2)x +1)))

$$\left.\mathrm{1}\right){let}\:\mathrm{0}<\theta<\frac{\pi}{\mathrm{2}}\:\:\:\:{and}\:\:{A}\left(\theta\right)\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{dx}}{\sqrt{{x}^{\mathrm{2}} \:+\mathrm{2}{sin}\theta\:{x}\:+\mathrm{1}}} \\ $$ $${calculate}\:{A}\left(\theta\right) \\ $$ $$\left.\mathrm{2}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{dx}}{\sqrt{{x}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}{x}\:+\mathrm{1}}} \\ $$

Commented bymaxmathsup 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)).

$$\left.\mathrm{1}\right)\:{we}\:{have}\:{A}\left(\theta\right)\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\frac{{dx}}{\sqrt{{x}^{\mathrm{2}} \:+\mathrm{2}{x}\:{sin}\theta\:+{sin}^{\mathrm{2}} \theta\:+{cos}^{\mathrm{2}} \theta}} \\ $$ $$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{dx}}{\sqrt{\left({x}+{sin}\theta\right)^{\mathrm{2}\:} +{cos}^{\mathrm{2}} \theta}}\:\:{changement}\:{x}+{sin}\theta={t}\:{cos}\theta\:{give}\:{t}\:=\frac{{x}+{sin}\theta}{{cos}\theta} \\ $$ $${A}\left(\theta\right)\:=\:\int_{{tan}\theta} ^{\frac{\pi}{\mathrm{2}{cos}\theta}+{tan}\theta} \:\:\:\:\:\frac{{cos}\theta\:{dt}}{{cos}\theta\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}\:=\:\int_{{tan}\theta} ^{\frac{\pi}{\mathrm{2}{cos}\theta}\:+{tan}\theta} \:\:\frac{{dt}}{\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}\:=\left[{argsh}\left({t}\right)\right]_{{tan}\theta} ^{\frac{\pi}{\mathrm{2}{cos}\theta}+{tan}\theta} \\ $$ $$={ln}\left({t}+\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\right]_{{tan}\theta} ^{\frac{\pi}{\mathrm{2}{cos}\theta}\:+{tan}\theta} \:=\:{ln}\left(\frac{\pi}{\mathrm{2}{cos}\theta}\:+{tan}\theta\:+\sqrt{\mathrm{1}+\left(\frac{\pi}{\mathrm{2}{cos}\theta}+{tan}\theta\right)^{\mathrm{2}} }\right) \\ $$ $$−{ln}\left({tan}\theta\:+\sqrt{\mathrm{1}+{tan}^{\mathrm{2}} \theta}\right) \\ $$ $$\left.\mathrm{2}\right)\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{dx}}{\sqrt{{x}^{\mathrm{2}} +\sqrt{\mathrm{2}}{x}\:+\mathrm{1}}}\:={A}\left(\frac{\pi}{\mathrm{4}}\right)\:={ln}\left(\frac{\pi}{\mathrm{2}\:{cos}\left(\frac{\pi}{\mathrm{4}}\right)}\:+{tan}\left(\frac{\pi}{\mathrm{4}}\right)+\sqrt{\mathrm{1}+\left(\frac{\pi}{\mathrm{2}{cos}\left(\frac{\pi}{\mathrm{4}}\right)}+{tan}\left(\frac{\pi}{\mathrm{4}}\right)^{\mathrm{2}} \right.}\right. \\ $$ $$−{ln}\left({tan}\left(\frac{\pi}{\mathrm{4}}\right)+\sqrt{\mathrm{1}+{tan}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{4}}\right)}\right) \\ $$ $$={ln}\left(\frac{\pi}{\sqrt{\mathrm{2}}}\:+\mathrm{1}\:+\sqrt{\mathrm{1}+\left(\frac{\pi}{\sqrt{\mathrm{2}}}+\mathrm{1}\right)^{\mathrm{2}} }\right)−{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right). \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 22/Jan/19

1)∫_0 ^(π/2) (dx/(√(x^2 +2xsinθ+1)))  ∫_0 ^(π/2) (dx/(√((x+sinθ)^2 +cos^2 θ)))=∣ln{∣(x+sinθ)+(√((x+sinθ)^2 +cos^2 θ)) )∣_0 ^(π/2)   =[ln{((π/2)+sinθ)+(√(((π/2)+sinθ)^2 +cos^2 θ)) }−ln{(0+sinθ)+(√((0+sinθ)^2 +cos^2 θ)) }  =ln{((π/2)+sinθ)+(√((π^2 /4)+πsinθ+1)) }−ln{sinθ+1}   using formula  ∫(dx/(√(x^2 +a^2 )))=ln(x+(√(x^2 +a^2 )) )  2)put θ=(π/4) for answer...  ln{((π/4)+(1/(√2)))+(√((π^2 /4)+(π/(√2))+1)) }−ln{(1/(√2))+1}  sir pls check...

$$\left.\mathrm{1}\right)\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{dx}}{\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{xsin}\theta+\mathrm{1}}} \\ $$ $$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{dx}}{\sqrt{\left({x}+{sin}\theta\right)^{\mathrm{2}} +{cos}^{\mathrm{2}} \theta}}=\mid{ln}\left\{\mid\left({x}+{sin}\theta\right)+\sqrt{\left({x}+{sin}\theta\right)^{\mathrm{2}} +{cos}^{\mathrm{2}} \theta}\:\right)\mid_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \\ $$ $$=\left[{ln}\left\{\left(\frac{\pi}{\mathrm{2}}+{sin}\theta\right)+\sqrt{\left(\frac{\pi}{\mathrm{2}}+{sin}\theta\right)^{\mathrm{2}} +{cos}^{\mathrm{2}} \theta}\:\right\}−{ln}\left\{\left(\mathrm{0}+{sin}\theta\right)+\sqrt{\left(\mathrm{0}+{sin}\theta\right)^{\mathrm{2}} +{cos}^{\mathrm{2}} \theta}\:\right\}\right. \\ $$ $$={ln}\left\{\left(\frac{\pi}{\mathrm{2}}+{sin}\theta\right)+\sqrt{\frac{\pi^{\mathrm{2}} }{\mathrm{4}}+\pi{sin}\theta+\mathrm{1}}\:\right\}−{ln}\left\{{sin}\theta+\mathrm{1}\right\} \\ $$ $$\:{using}\:{formula} \\ $$ $$\int\frac{{dx}}{\sqrt{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }}={ln}\left({x}+\sqrt{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }\:\right) \\ $$ $$\left.\mathrm{2}\right){put}\:\theta=\frac{\pi}{\mathrm{4}}\:{for}\:{answer}... \\ $$ $${ln}\left\{\left(\frac{\pi}{\mathrm{4}}+\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\right)+\sqrt{\frac{\pi^{\mathrm{2}} }{\mathrm{4}}+\frac{\pi}{\sqrt{\mathrm{2}}}+\mathrm{1}}\:\right\}−{ln}\left\{\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}+\mathrm{1}\right\} \\ $$ $${sir}\:{pls}\:{check}... \\ $$

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