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Question Number 182726 by SANOGO last updated on 13/Dec/22

$$\int_{\mathrm{0}} ^{\mathrm{1}} \:{t}\sqrt{\frac{\mathrm{1}−{t}}{\mathrm{1}+{t}}}{dt}=? \\$$

Commented by CElcedricjunior last updated on 13/Dec/22

$$\int_{\mathrm{0}} ^{\mathrm{1}} \boldsymbol{{t}}\sqrt{\frac{\mathrm{1}−\boldsymbol{{t}}}{\mathrm{1}+\boldsymbol{{t}}}}\boldsymbol{{dt}}=\boldsymbol{{k}} \\$$$$\boldsymbol{{posons}}\:\sqrt{\frac{\mathrm{1}−\boldsymbol{{t}}}{\mathrm{1}+\boldsymbol{{t}}}}=\boldsymbol{{x}} \\$$$$\frac{\mathrm{1}−\boldsymbol{{t}}}{\mathrm{1}+\boldsymbol{{t}}}=\boldsymbol{{x}}^{\mathrm{2}} =>\mathrm{1}−\boldsymbol{{t}}=\boldsymbol{{tx}}^{\mathrm{2}} +\boldsymbol{{x}}^{\mathrm{2}} =>\boldsymbol{{t}}=\frac{\mathrm{1}−\boldsymbol{{x}}^{\mathrm{2}} }{\mathrm{1}+\boldsymbol{{x}}^{\mathrm{2}} } \\$$$$\boldsymbol{{dt}}=\frac{−\mathrm{4}\boldsymbol{{x}}}{\left(\mathrm{1}+\boldsymbol{{x}}^{\mathrm{2}} \right)^{\mathrm{2}} }\boldsymbol{{dx}}=>\boldsymbol{{qd}}:\begin{cases}{\boldsymbol{{t}}=\mathrm{0}}\\{\boldsymbol{{t}}=\mathrm{1}}\end{cases}=>\begin{cases}{\boldsymbol{{x}}=\mathrm{1}}\\{\boldsymbol{{x}}=\mathrm{0}}\end{cases} \\$$$$\boldsymbol{{k}}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{4}\boldsymbol{{x}}\left(\mathrm{1}−\boldsymbol{{x}}^{\mathrm{2}} \right)}{\left(\mathrm{1}+\boldsymbol{{x}}^{\mathrm{2}} \right)^{\mathrm{3}} }\boldsymbol{{dx}}=\int\frac{−\mathrm{4}\boldsymbol{{x}}\left(\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{1}\right)+\mathrm{8}\boldsymbol{{x}}}{\left(\mathrm{1}+\boldsymbol{{x}}^{\mathrm{2}} \right)^{\mathrm{3}} }\boldsymbol{{dx}} \\$$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{−\mathrm{4}\boldsymbol{{x}}}{\left(\mathrm{1}+\boldsymbol{{x}}^{\mathrm{2}} \underset{} {\right)}^{\mathrm{2}} }\boldsymbol{{dx}}+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{8}\boldsymbol{{x}}}{\left(\mathrm{1}+\boldsymbol{{x}}^{\mathrm{2}} \right)^{\mathrm{3}} }\boldsymbol{{dx}} \\$$$$=\left[\frac{\mathrm{2}}{\mathrm{1}+\boldsymbol{{x}}^{\mathrm{2}} }\right]_{\mathrm{0}} ^{\mathrm{1}} −\left[\frac{\mathrm{2}}{\left(\mathrm{1}+\boldsymbol{{x}}^{\mathrm{2}} \right)^{\mathrm{2}} }\right]_{\mathrm{0}} ^{\mathrm{1}} \\$$$$=\mathrm{1}−\mathrm{2}−\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{2} \\$$$$\boldsymbol{{k}}=\frac{\mathrm{1}}{\mathrm{2}} \\$$$$======================= \\$$$$..........{le}\:{celebre}\:{cedric}\:{junior}........\mathrm{673804515} \\$$$$==================================== \\$$$$\\$$

Commented by ARUNG_Brandon_MBU last updated on 13/Dec/22

$${k}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{−\mathrm{4}{x}^{\mathrm{2}} \left(\mathrm{1}−{x}^{\mathrm{2}} \right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{3}} }{dx} \\$$

Commented by ARUNG_Brandon_MBU last updated on 13/Dec/22

Commented by SANOGO last updated on 13/Dec/22

$${merci} \\$$

Answered by SEKRET last updated on 13/Dec/22

$$\int_{\mathrm{0}} ^{\mathrm{1}} \boldsymbol{\mathrm{t}}\centerdot\frac{\sqrt{\mathrm{1}−\boldsymbol{\mathrm{t}}^{\mathrm{2}} }}{\mathrm{1}+\boldsymbol{\mathrm{t}}}\:\boldsymbol{\mathrm{dt}}\:\:\:\begin{vmatrix}{\boldsymbol{\mathrm{sin}}\left(\boldsymbol{\mathrm{t}}\right)=\:\boldsymbol{\mathrm{x}}}\\{\boldsymbol{\mathrm{dx}}=\boldsymbol{\mathrm{cos}}\left(\boldsymbol{\mathrm{t}}\right)\:\boldsymbol{\mathrm{dt}}}\end{vmatrix} \\$$$$\:\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \boldsymbol{\mathrm{sin}}\left(\boldsymbol{\mathrm{t}}\right)\centerdot\frac{\boldsymbol{\mathrm{cos}}\left(\boldsymbol{\mathrm{t}}\right)}{\mathrm{1}+\boldsymbol{\mathrm{sin}}\left(\boldsymbol{\mathrm{t}}\right)}\centerdot\boldsymbol{\mathrm{cos}}\left(\boldsymbol{\mathrm{t}}\right)\:\boldsymbol{\mathrm{dt}}= \\$$$$\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \frac{\boldsymbol{\mathrm{sin}}\left(\boldsymbol{\mathrm{t}}\right)\centerdot\left(\mathrm{1}−\boldsymbol{\mathrm{sint}}\right)\left(\mathrm{1}+\boldsymbol{\mathrm{sint}}\right)}{\mathrm{1}+\boldsymbol{\mathrm{sint}}}\boldsymbol{\mathrm{dt}}= \\$$$$\int_{\mathrm{0}} ^{\:\frac{\boldsymbol{\pi}}{\mathrm{2}}} \boldsymbol{\mathrm{sin}}\left(\boldsymbol{\mathrm{t}}\right)\:−\:\boldsymbol{\mathrm{sin}}^{\mathrm{2}} \left(\boldsymbol{\mathrm{t}}\right)\:\boldsymbol{\mathrm{dt}}= \\$$$$\:\int_{\mathrm{0}} ^{\:\frac{\boldsymbol{\pi}}{\mathrm{2}}} \boldsymbol{\mathrm{sin}}\left(\boldsymbol{\mathrm{t}}\right)\boldsymbol{\mathrm{dt}}−\:\frac{\mathrm{1}}{\mathrm{2}}\centerdot\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \mathrm{1}−\boldsymbol{\mathrm{cos}}\left(\mathrm{2}\boldsymbol{\mathrm{t}}\right)\:\boldsymbol{\mathrm{dt}} \\$$$$\:\:\:−\boldsymbol{\mathrm{cos}}\left(\boldsymbol{\mathrm{t}}\right)\:/_{\mathrm{0}} ^{\:\frac{\boldsymbol{\pi}}{\mathrm{2}}} −\frac{\mathrm{1}}{\mathrm{2}}\centerdot\boldsymbol{\mathrm{t}}+\frac{\mathrm{1}}{\mathrm{4}}\centerdot\boldsymbol{\mathrm{sin}}\left(\mathrm{2}\boldsymbol{\mathrm{t}}\right)\:/_{\mathrm{0}} ^{\:\frac{\boldsymbol{\pi}}{\mathrm{2}}} = \\$$$$\:\:\:\:−\left(\mathrm{0}−\mathrm{1}\right)−\frac{\mathrm{1}}{\mathrm{2}}\centerdot\frac{\boldsymbol{\pi}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}\centerdot\mathrm{0}=\mathrm{1}−\frac{\boldsymbol{\pi}}{\mathrm{4}} \\$$$$\:\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \boldsymbol{\mathrm{t}}\centerdot\sqrt{\frac{\mathrm{1}−\boldsymbol{\mathrm{t}}}{\mathrm{1}+\boldsymbol{\mathrm{t}}}\:}\:\boldsymbol{\mathrm{dt}}\:\:=\:\:\mathrm{1}\:\:−\:\:\frac{\boldsymbol{\pi}}{\mathrm{4}} \\$$$$\:\boldsymbol{{ABDULAZIZ}}\:\:\:\boldsymbol{{ABDUVALIYEV}} \\$$

Answered by ARUNG_Brandon_MBU last updated on 13/Dec/22

$$\int_{\mathrm{0}} ^{\mathrm{1}} {t}\sqrt{\frac{\mathrm{1}−{t}}{\mathrm{1}+{t}}}{dt}=\int_{\mathrm{0}} ^{\mathrm{1}} {t}\centerdot\frac{\mathrm{1}−{t}}{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{dt} \\$$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}}{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{\mathrm{2}} }{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{dt} \\$$$$=−\left[\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\right]_{\mathrm{0}} ^{\mathrm{1}} −\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{sin}^{\mathrm{2}} \theta{d}\theta \\$$$$=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\mathrm{1}−\mathrm{cos2}\theta\right){d}\theta=\mathrm{1}−\frac{\pi}{\mathrm{4}} \\$$