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Question Number 54033 by qw last updated on 28/Jan/19

$$\underset{\:\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\:\sqrt{\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}}\:{dx}\:= \\$$

Commented by maxmathsup by imad last updated on 29/Jan/19

$${let}\:{I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}}{dx}\:\:{let}\:{use}\:{the}\:{chan}.\:\sqrt{\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}}={t}\:\Rightarrow\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\:={t}^{\mathrm{2}} \:\Rightarrow \\$$$$\mathrm{1}−{x}\:={t}^{\mathrm{2}} \:+{t}^{\mathrm{2}} {x}\:\Rightarrow\mathrm{1}−{t}^{\mathrm{2}} =\left(\mathrm{1}+{t}^{\mathrm{2}} \right){x}\:\:\Rightarrow{x}\:=\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }\:\Rightarrow\:\frac{{dx}}{{dt}}\:=\frac{−\mathrm{2}{t}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)−\left(\mathrm{1}−{t}^{\mathrm{2}} \right)\left(\mathrm{2}{t}\right)}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} } \\$$$$=\frac{−\mathrm{2}{t}−\mathrm{2}{t}^{\mathrm{3}} −\mathrm{2}{t}\:+\mathrm{2}{t}^{\mathrm{3}} }{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\:=\frac{−\mathrm{4}{t}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\:\Rightarrow\:{I}\:=\:−\int_{\mathrm{0}} ^{\mathrm{1}} \:{t}\left(\frac{−\mathrm{4}{t}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\:\right){dt} \\$$$$=\mathrm{4}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{t}^{\mathrm{2}} }{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\:{dt}\:=\mathrm{4}\left\{\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{1}+{t}^{\mathrm{2}} −\mathrm{1}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{dt}\right\}=\mathrm{4}\left\{\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }\:−\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\right\}\:{but} \\$$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }\:=\left[{arctant}\right]_{\mathrm{0}} ^{\mathrm{1}} =\frac{\pi}{\mathrm{4}} \\$$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\:=_{{t}\:={tan}\theta} \:\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\:\frac{\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right)}{\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right)^{\mathrm{2}} }{d}\theta\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\frac{{d}\theta}{\mathrm{1}+{tan}^{\mathrm{2}} \theta} \\$$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{cos}^{\mathrm{2}} \theta\:{d}\theta\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{\mathrm{1}+{cos}\left(\mathrm{2}\theta\right)}{\mathrm{2}}\:{d}\theta\:=\frac{\pi}{\mathrm{8}}\:+\frac{\mathrm{1}}{\mathrm{4}}\left[{sin}\left(\mathrm{2}\theta\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} =\frac{\pi}{\mathrm{8}}\:+\frac{\mathrm{1}}{\mathrm{4}}\:\Rightarrow \\$$$${I}\:=\mathrm{4}\left\{\frac{\pi}{\mathrm{4}}\:−\frac{\pi}{\mathrm{8}}\:−\frac{\mathrm{1}}{\mathrm{4}}\right\}\:=\pi−\frac{\pi}{\mathrm{2}}\:−\mathrm{1}\:=\frac{\pi}{\mathrm{2}}\:−\mathrm{1}\:. \\$$

Answered by tanmay.chaudhury50@gmail.com last updated on 28/Jan/19

$$\int\frac{\sqrt{\mathrm{1}−{x}}}{\sqrt{\mathrm{1}+{x}}}{dx} \\$$$$\int\frac{\mathrm{1}−{x}}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx} \\$$$$\int\frac{{dx}}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{d}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} \\$$$${sin}^{−\mathrm{1}} \left({x}\right)+\frac{\mathrm{1}}{\mathrm{2}}×\frac{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\frac{−\mathrm{1}}{\mathrm{2}}+\mathrm{1}} }{\frac{\mathrm{1}}{\mathrm{2}}}+{c} \\$$$${so}\:{required}\:{answer}\:{is} \\$$$$\mid{sin}^{−\mathrm{1}} {x}+\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} \mid_{\mathrm{0}} ^{\mathrm{1}} \\$$$$={sin}^{−\mathrm{1}} \left(\mathrm{1}\right)+\mathrm{0}−{sin}^{−\mathrm{1}} \left(\mathrm{0}\right)−\left(\mathrm{1}−\mathrm{0}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \\$$$$=\frac{\pi}{\mathrm{2}}−\mathrm{1} \\$$

Answered by ajfour last updated on 28/Jan/19

$${let}\:{x}=\mathrm{cos}\:\mathrm{2}\theta \\$$$$\int_{\pi/\mathrm{4}} ^{\:\:\mathrm{0}} \mathrm{tan}\:\theta\left(−\mathrm{4sin}\:\theta\mathrm{cos}\:\theta\right){d}\theta \\$$$$=\mathrm{2}\int_{\mathrm{0}} ^{\:\:\pi/\mathrm{4}} \left(\mathrm{1}−\mathrm{cos}\:\mathrm{2}\theta\right){d}\theta\: \\$$$$\:=\:\frac{\pi}{\mathrm{2}}−\mathrm{1}\:. \\$$