Question Number 42435 by maxmathsup by imad last updated on 25/Aug/18
$${find}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{dx}}{\:\sqrt{{x}}\:+\sqrt{\mathrm{1}−{x}}} \\ $$
Commented by maxmathsup by imad last updated on 25/Aug/18
![let I = ∫_0 ^1 (dx/( (√x) +(√(1−x)))) changement x=sin^2 t give I = ∫_0 ^(π/2) ((2 sint cost dt)/(sint +cost)) = ∫_0 ^(π/2) ((sin^2 t +cos^2 t +2sint cost −1)/(sint +cost))dt = ∫_0 ^(π/2) (((sint +cost)^2 −1)/(sint +cost)) dt =∫_0 ^(π/2) (sint +cost)dt −∫_0 ^(π/2) (dt/(sint +cost)) =[sint −cost]_0 ^(π/2) − ∫_0 ^(π/2) (dt/(sint +cost)) =2 −∫_0 ^(π/2) (dt/(sint +cost)) changement tan((t/2))=u give ∫_0 ^(π/2) (dt/(sint +cost)) =∫_0 ^1 (1/(((2u)/(1+u^2 )) +((1−u^2 )/(1+u^2 )))) ((2du)/(1+u^2 )) = 2 ∫_0 ^1 (du/(2u+1−u^2 )) =−2 ∫_0 ^1 (du/(u^2 −2u −1)) =−2 ∫_0 ^1 (du/((u−1)^2 −1)) =_(u−1 =−α) −2 ∫_1 ^0 ((−dα)/(α^2 −1)) =−2 ∫_0 ^1 (dα/(α^2 −1)) =− ∫_0 ^1 ((1/(α−1))−(1/(α+1))) =−[ln∣((α−1)/(α +1))∣_0 ^1 = +∞ so this integral diverges !..](https://www.tinkutara.com/question/Q42460.png)
$${let}\:{I}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{{dx}}{\:\sqrt{{x}}\:+\sqrt{\mathrm{1}−{x}}}\:\:\:{changement}\:{x}={sin}^{\mathrm{2}} {t}\:{give} \\ $$$${I}\:=\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\frac{\mathrm{2}\:{sint}\:{cost}\:{dt}}{{sint}\:+{cost}}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{sin}^{\mathrm{2}} {t}\:+{cos}^{\mathrm{2}} {t}\:+\mathrm{2}{sint}\:{cost}\:−\mathrm{1}}{{sint}\:+{cost}}{dt} \\ $$$$=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{\left({sint}\:+{cost}\right)^{\mathrm{2}} −\mathrm{1}}{{sint}\:+{cost}}\:{dt}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left({sint}\:+{cost}\right){dt}\:−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{dt}}{{sint}\:+{cost}} \\ $$$$=\left[{sint}\:−{cost}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:−\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{dt}}{{sint}\:+{cost}} \\ $$$$=\mathrm{2}\:−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{dt}}{{sint}\:+{cost}}\:\:{changement}\:{tan}\left(\frac{{t}}{\mathrm{2}}\right)={u}\:{give} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{dt}}{{sint}\:+{cost}}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\:\frac{\mathrm{1}}{\frac{\mathrm{2}{u}}{\mathrm{1}+{u}^{\mathrm{2}} }\:+\frac{\mathrm{1}−{u}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{2}} }}\:\:\frac{\mathrm{2}{du}}{\mathrm{1}+{u}^{\mathrm{2}} }\:=\:\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{{du}}{\mathrm{2}{u}+\mathrm{1}−{u}^{\mathrm{2}} } \\ $$$$=−\mathrm{2}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{du}}{{u}^{\mathrm{2}} −\mathrm{2}{u}\:−\mathrm{1}}\:\:=−\mathrm{2}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{{du}}{\left({u}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1}}\:=_{{u}−\mathrm{1}\:=−\alpha} \:−\mathrm{2}\:\:\int_{\mathrm{1}} ^{\mathrm{0}} \:\:\:\:\frac{−{d}\alpha}{\alpha^{\mathrm{2}} −\mathrm{1}}\:\:\:\: \\ $$$$=−\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{d}\alpha}{\alpha^{\mathrm{2}} −\mathrm{1}}\:=−\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\left(\frac{\mathrm{1}}{\alpha−\mathrm{1}}−\frac{\mathrm{1}}{\alpha+\mathrm{1}}\right)\:=−\left[{ln}\mid\frac{\alpha−\mathrm{1}}{\alpha\:+\mathrm{1}}\mid_{\mathrm{0}} ^{\mathrm{1}} \:=\:+\infty\:{so}\:{this}\right. \\ $$$${integral}\:{diverges}\:!.. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 25/Aug/18
$${I}= \\ $$$${x}={sin}^{\mathrm{2}} \alpha\:\:\:{dx}=\mathrm{2}{sin}\alpha{cos}\alpha \\ $$$$\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} \frac{\mathrm{2}{sin}\alpha{cos}\alpha}{{sin}\alpha+{cos}\alpha}{d}\alpha \\ $$$$\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} \frac{{sin}^{\mathrm{2}} \alpha+{cos}^{\mathrm{2}} \alpha+\mathrm{2}{sin}\alpha{cos}\alpha−\mathrm{1}}{{sin}\alpha+{cos}\alpha}{d}\alpha \\ $$$$\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} {sin}\alpha+{cos}\alpha\:{d}\alpha−\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} \frac{{d}\alpha}{{sin}\alpha+{cos}\alpha} \\ $$$$\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} {sin}\alpha+{cos}\alpha\:{d}\alpha−\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} \frac{{sec}^{\mathrm{2}} \frac{\alpha}{\mathrm{2}}}{\mathrm{2}{tan}\frac{\alpha}{\mathrm{2}}+\mathrm{1}−{tan}^{\mathrm{2}} \frac{\alpha}{\mathrm{2}}}{d}\alpha \\ $$$$\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} {sin}\alpha+{cos}\alpha\:{d}\alpha+\mathrm{2}\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} \frac{{d}\left({tan}\frac{\alpha}{\mathrm{2}}\right)}{{tan}^{\mathrm{2}} \frac{\alpha}{\mathrm{2}}−\mathrm{2}{tan}\frac{\alpha}{\mathrm{2}}+\mathrm{1}−\mathrm{2}} \\ $$$$\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} {sin}\alpha+{cos}\alpha\:{d}\alpha+\mathrm{2}\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} \frac{{d}\left({tan}\frac{\alpha}{\mathrm{2}}\right)}{\left({tan}\frac{\alpha}{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1}} \\ $$$$\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} {sin}\alpha+{cos}\alpha\:{d}\alpha−\mathrm{2}\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} \frac{{d}\left({tan}\frac{\alpha}{\mathrm{2}}\right)}{\mathrm{1}−\left({tan}\frac{\alpha}{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} \:{sin}\alpha+{cos}\alpha\:{d}\alpha−\mathrm{2}\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} \frac{{d}\left({tan}\frac{\alpha}{\mathrm{2}}\:−\mathrm{1}\right)}{\left({tan}\frac{\alpha}{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1}}\rightarrow\int\frac{{dx}}{{x}^{\mathrm{2}} −{a}^{\mathrm{2}} } \\ $$$$=\mid−{cos}\alpha+{sin}\alpha\mid_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} −\mathrm{2}×\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid\frac{{tan}\frac{\alpha}{\mathrm{2}}−\mathrm{1}−\mathrm{1}}{{tan}\frac{\alpha}{\mathrm{2}}−\mathrm{1}+\mathrm{1}}\mid_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} \\ $$$$=\left\{\left(−\mathrm{0}+\mathrm{1}\right)−\left(−\mathrm{1}+\mathrm{0}\right)\right\}−\left\{{ln}\mid\frac{\mathrm{1}−\mathrm{1}−\mathrm{1}}{\mathrm{1}−\mathrm{1}+\mathrm{1}}\mid−{ln}\mid\frac{−\mathrm{2}}{\mathrm{0}}\mid\right\} \\ $$$$=\mathrm{2}−\left\{{ln}\mid−\mathrm{1}\mid−{ln}\mid−\infty\mid\right\} \\ $$$$ \\ $$$$ \\ $$