Question Number 162189 by SANOGO last updated on 27/Dec/21
$${show}\:{the}\:{converge}^{} {nce}\:{and}\:{calculate} \\ $$$$\int_{−\mathrm{1}} ^{\mathrm{1}} \sqrt{\frac{\mathrm{1}−{t}}{\mathrm{1}+{t}}}\:{dt} \\ $$
Commented by Ar Brandon last updated on 27/Dec/21
![I=∫_(−1) ^1 (√((1−t)/(1+t)))dt=∫_(−1) ^1 ((1−t)/( (√(1−t^2 ))))dt =∫_(−1) ^1 (dt/( (√(1−t^2 ))))−∫(t/( (√(1−t^2 ))))dt =[arcsin(t)]_(−1) ^1 −[(√(1−t^2 ))]_(−1) ^1 =π](Q162221.png)
$${I}=\int_{−\mathrm{1}} ^{\mathrm{1}} \sqrt{\frac{\mathrm{1}−{t}}{\mathrm{1}+{t}}}{dt}=\int_{−\mathrm{1}} ^{\mathrm{1}} \frac{\mathrm{1}−{t}}{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{dt} \\ $$$$\:\:\:=\int_{−\mathrm{1}} ^{\mathrm{1}} \frac{{dt}}{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}−\int\frac{{t}}{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{dt} \\ $$$$\:\:\:=\left[\mathrm{arcsin}\left({t}\right)\right]_{−\mathrm{1}} ^{\mathrm{1}} −\left[\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\right]_{−\mathrm{1}} ^{\mathrm{1}} =\pi \\ $$
Commented by SANOGO last updated on 28/Dec/21
$${thank}\:{you} \\ $$
Answered by mathmax by abdo last updated on 27/Dec/21
![I=∫_(−1) ^1 (√((1−t)/(1+t)))dt ⇒I=∫_(−1) ^0 (√((1−t)/(1+t)))dt +∫_0 ^1 (√((1−t)/(1+t)))dt ∫_(−1) ^0 (√((1−t)/(1+t)))dt =_(t=−u) −∫_0 ^1 (√((1+u)/(1−u)))du =_(u=cosθ) −∫_(π/2) ^0 (√((2cos^2 ((θ/2)))/(2sin^2 ((θ/2)))))(−sinθ)dθ =−∫_0 ^(π/2) ((cos((θ/2)))/(sin((θ/2))))×2sin((θ/2))cos((θ/2))dθ =−2∫_0 ^(π/2) cos^2 ((θ/2))dθ =−∫_0 ^(π/2) (1+cosθ)dθ =−(π/2)−[sinθ]_0 ^(π/2) =−(π/2)−1 ∫_0 ^1 (√((1−t)/(1+t)))dt =_(t=cosθ) ∫_(π/2) ^0 (√((1−cosθ)/(1+cosθ)))(−sinθ)dθ =∫_0 ^(π/2) (√((2sin^2 ((θ/2)))/(2cos^2 ((θ/2)))))sinθ dθ =∫_0 ^(π/2) ((sin((θ/2)))/(cos((θ/2))))×2sin((θ/2))cos((θ/2))dθ =2∫_0 ^(π/2) sin^2 ((θ/2))dθ =∫_0 ^(π/2) (1−cos(θ))dθ =(π/2)−[sinθ]_0 ^(π/2) =(π/2)−1 ⇒ I=−(π/2)−1+(π/2)−1 ⇒I =−2](Q162201.png)
$$\mathrm{I}=\int_{−\mathrm{1}} ^{\mathrm{1}} \sqrt{\frac{\mathrm{1}−\mathrm{t}}{\mathrm{1}+\mathrm{t}}}\mathrm{dt}\:\:\Rightarrow\mathrm{I}=\int_{−\mathrm{1}} ^{\mathrm{0}} \sqrt{\frac{\mathrm{1}−\mathrm{t}}{\mathrm{1}+\mathrm{t}}}\mathrm{dt}\:+\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\frac{\mathrm{1}−\mathrm{t}}{\mathrm{1}+\mathrm{t}}}\mathrm{dt} \\ $$$$\int_{−\mathrm{1}} ^{\mathrm{0}} \sqrt{\frac{\mathrm{1}−\mathrm{t}}{\mathrm{1}+\mathrm{t}}}\mathrm{dt}\:=_{\mathrm{t}=−\mathrm{u}} \:\:\:−\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\frac{\mathrm{1}+\mathrm{u}}{\mathrm{1}−\mathrm{u}}}\mathrm{du} \\ $$$$=_{\mathrm{u}=\mathrm{cos}\theta} \:\:\:−\int_{\frac{\pi}{\mathrm{2}}} ^{\mathrm{0}} \sqrt{\frac{\mathrm{2cos}^{\mathrm{2}} \left(\frac{\theta}{\mathrm{2}}\right)}{\mathrm{2sin}^{\mathrm{2}} \left(\frac{\theta}{\mathrm{2}}\right)}}\left(−\mathrm{sin}\theta\right)\mathrm{d}\theta \\ $$$$=−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{\mathrm{cos}\left(\frac{\theta}{\mathrm{2}}\right)}{\mathrm{sin}\left(\frac{\theta}{\mathrm{2}}\right)}×\mathrm{2sin}\left(\frac{\theta}{\mathrm{2}}\right)\mathrm{cos}\left(\frac{\theta}{\mathrm{2}}\right)\mathrm{d}\theta \\ $$$$=−\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{cos}^{\mathrm{2}} \left(\frac{\theta}{\mathrm{2}}\right)\mathrm{d}\theta\:=−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\mathrm{1}+\mathrm{cos}\theta\right)\mathrm{d}\theta \\ $$$$=−\frac{\pi}{\mathrm{2}}−\left[\mathrm{sin}\theta\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:=−\frac{\pi}{\mathrm{2}}−\mathrm{1} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\frac{\mathrm{1}−\mathrm{t}}{\mathrm{1}+\mathrm{t}}}\mathrm{dt}\:=_{\mathrm{t}=\mathrm{cos}\theta} \:\:\:\int_{\frac{\pi}{\mathrm{2}}} ^{\mathrm{0}} \sqrt{\frac{\mathrm{1}−\mathrm{cos}\theta}{\mathrm{1}+\mathrm{cos}\theta}}\left(−\mathrm{sin}\theta\right)\mathrm{d}\theta \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \sqrt{\frac{\mathrm{2sin}^{\mathrm{2}} \left(\frac{\theta}{\mathrm{2}}\right)}{\mathrm{2cos}^{\mathrm{2}} \left(\frac{\theta}{\mathrm{2}}\right)}}\mathrm{sin}\theta\:\mathrm{d}\theta \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{sin}\left(\frac{\theta}{\mathrm{2}}\right)}{\mathrm{cos}\left(\frac{\theta}{\mathrm{2}}\right)}×\mathrm{2sin}\left(\frac{\theta}{\mathrm{2}}\right)\mathrm{cos}\left(\frac{\theta}{\mathrm{2}}\right)\mathrm{d}\theta \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{sin}^{\mathrm{2}} \left(\frac{\theta}{\mathrm{2}}\right)\mathrm{d}\theta\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\mathrm{1}−\mathrm{cos}\left(\theta\right)\right)\mathrm{d}\theta \\ $$$$=\frac{\pi}{\mathrm{2}}−\left[\mathrm{sin}\theta\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:=\frac{\pi}{\mathrm{2}}−\mathrm{1}\:\Rightarrow \\ $$$$\mathrm{I}=−\frac{\pi}{\mathrm{2}}−\mathrm{1}+\frac{\pi}{\mathrm{2}}−\mathrm{1}\:\Rightarrow\mathrm{I}\:=−\mathrm{2} \\ $$
Commented by SANOGO last updated on 27/Dec/21
$${thank}\:{you} \\ $$