Question Number 42394 by abdo.msup.com last updated on 24/Aug/18
$${find}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{dt}}{{t}+\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}\:{dt} \\ $$
Commented by maxmathsup by imad last updated on 25/Aug/18
![changement t = sinx give I = ∫_0 ^(π/2) ((cosxdx)/(sinx +cosx)) dx = ∫_0 ^(π/2) (dx/(tanx +1)) I =_(tanx =u) ∫_0 ^(+∞) (1/(1+u)) (du/(1+u^2 )) = ∫_0 ^∞ (du/((u+1)(u^2 +1))) let decompose F(u) = (1/((u+1)(u^2 +1))) ⇒F(u) =(a/(u+1)) +((bu +c)/(u^2 +1)) a =lim_(u→−1) (u+1)F(u) =(1/2) lim_(u→+∞) u F(u) =0 =a+b ⇒b =−(1/2) ⇒F(u) =(1/(2(u+1))) +((−(1/2)u +c)/(u^2 +1)) F(0) =1 =(1/2) +c ⇒c =(1/2) ⇒F(u) = (1/(2(u+1))) −(1/2) ((u−1)/(u^2 +1)) ⇒ I = ∫_0 ^∞ F(u)du = (1/2) ∫_0 ^∞ (du/(u+1)) −(1/4) ∫_0 ^∞ ((2u−2)/(u^2 +1))du =[ (1/2)ln∣u+1∣−(1/4)ln(u^2 +1)]_0 ^(+∞) +(1/2) [arctanu]_0 ^(+∞) =(1/2)[ln∣ ((u+1)/( (√(u^2 +1))))∣]_0 ^(+∞) +(π/4) =0+(π/4) =(π/4) .](https://www.tinkutara.com/question/Q42404.png)
$${changement}\:\:\:{t}\:=\:{sinx}\:{give} \\ $$$${I}\:\:=\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\:\frac{{cosxdx}}{{sinx}\:+{cosx}}\:{dx}\:\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\:\:\frac{{dx}}{{tanx}\:+\mathrm{1}} \\ $$$${I}\:=_{{tanx}\:={u}} \:\:\:\int_{\mathrm{0}} ^{+\infty} \:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{1}+{u}}\:\:\frac{{du}}{\mathrm{1}+{u}^{\mathrm{2}} }\:\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\:\:\frac{{du}}{\left({u}+\mathrm{1}\right)\left({u}^{\mathrm{2}} \:+\mathrm{1}\right)}\:\:{let}\:{decompose} \\ $$$${F}\left({u}\right)\:=\:\frac{\mathrm{1}}{\left({u}+\mathrm{1}\right)\left({u}^{\mathrm{2}} +\mathrm{1}\right)}\:\Rightarrow{F}\left({u}\right)\:=\frac{{a}}{{u}+\mathrm{1}}\:+\frac{{bu}\:+{c}}{{u}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${a}\:={lim}_{{u}\rightarrow−\mathrm{1}} \left({u}+\mathrm{1}\right){F}\left({u}\right)\:=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${lim}_{{u}\rightarrow+\infty} {u}\:{F}\left({u}\right)\:=\mathrm{0}\:={a}+{b}\:\Rightarrow{b}\:=−\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow{F}\left({u}\right)\:=\frac{\mathrm{1}}{\mathrm{2}\left({u}+\mathrm{1}\right)}\:\:+\frac{−\frac{\mathrm{1}}{\mathrm{2}}{u}\:+{c}}{{u}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${F}\left(\mathrm{0}\right)\:=\mathrm{1}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\:+{c}\:\Rightarrow{c}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow{F}\left({u}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}\left({u}+\mathrm{1}\right)}\:\:−\frac{\mathrm{1}}{\mathrm{2}}\:\frac{{u}−\mathrm{1}}{{u}^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow \\ $$$${I}\:=\:\int_{\mathrm{0}} ^{\infty} \:{F}\left({u}\right){du}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{du}}{{u}+\mathrm{1}}\:\:−\frac{\mathrm{1}}{\mathrm{4}}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{2}{u}−\mathrm{2}}{{u}^{\mathrm{2}} \:+\mathrm{1}}{du} \\ $$$$=\left[\:\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid{u}+\mathrm{1}\mid−\frac{\mathrm{1}}{\mathrm{4}}{ln}\left({u}^{\mathrm{2}} \:+\mathrm{1}\right)\right]_{\mathrm{0}} ^{+\infty} \:\:+\frac{\mathrm{1}}{\mathrm{2}}\:\left[{arctanu}\right]_{\mathrm{0}} ^{+\infty} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[{ln}\mid\:\frac{{u}+\mathrm{1}}{\:\sqrt{{u}^{\mathrm{2}} \:+\mathrm{1}}}\mid\right]_{\mathrm{0}} ^{+\infty} \:\:+\frac{\pi}{\mathrm{4}}\:\:=\mathrm{0}+\frac{\pi}{\mathrm{4}}\:=\frac{\pi}{\mathrm{4}}\:. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 24/Aug/18
$${t}={sin}\alpha\:\:{dt}={cos}\alpha{d}\alpha \\ $$$$\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} \frac{{cos}\alpha}{{cos}\alpha+{sin}\alpha}{d}\alpha \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} \frac{{cos}\alpha+{sin}\alpha+{cos}\alpha−{sin}\alpha}{{cos}\alpha+{sin}\alpha}{d}\alpha \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} {d}\alpha+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} \frac{{d}\left({sin}\alpha+{cos}\alpha\right)}{{cos}\alpha+{sin}\alpha} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\Pi}{\mathrm{2}}\right)+\frac{\mathrm{1}}{\mathrm{2}}\mid{ln}\left({cos}\alpha+{sin}\alpha\right)\mid_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} \\ $$$$=\frac{\Pi}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}\left\{{ln}\left(\mathrm{1}\right)−{ln}\mathrm{1}\right\} \\ $$$$=\frac{\Pi}{\mathrm{4}} \\ $$