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Question Number 162189 by SANOGO last updated on 27/Dec/21

$$showthe{converge}^{}nceandcalculate$$$${\int }_{-1}^1\sqrt{\frac{1-t}{1+t}}dt$$

Commented by Ar Brandon last updated on 27/Dec/21

$$I={\int }_{-1}^1\sqrt{\frac{1-t}{1+t}}dt={\int }_{-1}^1\frac{1-t}{\sqrt{1-t^2}}dt$$$$={\int }_{-1}^1\frac{dt}{\sqrt{1-t^2}}-\int \frac{t}{\sqrt{1-t^2}}dt$$$$={\left[\arcsin \left(t\right)\right]}_{-1}^1-{\left[\sqrt{1-t^2}\right]}_{-1}^1=\pi$$

Commented by SANOGO last updated on 28/Dec/21

$$thankyou$$

Answered by mathmax by abdo last updated on 27/Dec/21

$$\mathrm{I}={\int }_{-1}^1\sqrt{\frac{1-\mathrm{t}}{1+\mathrm{t}}}\mathrm{dt}\Rightarrow \mathrm{I}={\int }_{-1}^0\sqrt{\frac{1-\mathrm{t}}{1+\mathrm{t}}}\mathrm{dt}+{\int }_0^1\sqrt{\frac{1-\mathrm{t}}{1+\mathrm{t}}}\mathrm{dt}$$$${\int }_{-1}^0\sqrt{\frac{1-\mathrm{t}}{1+\mathrm{t}}}\mathrm{dt}{=}_{\mathrm{t}=-\mathrm{u}}-{\int }_0^1\sqrt{\frac{1+\mathrm{u}}{1-\mathrm{u}}}\mathrm{du}$$$${=}_{\mathrm{u}=\cos \theta }-{\int }_{\frac{\pi }{2}}^0\sqrt{\frac{{2\cos }^2\left(\frac{\theta }{2}\right)}{{2\sin }^2\left(\frac{\theta }{2}\right)}}(-\sin \theta )\mathrm{d}\theta$$$$=-{\int }_0^{\frac{\pi }{2}}\frac{\cos \left(\frac{\theta }{2}\right)}{\sin \left(\frac{\theta }{2}\right)}\times 2\sin \left(\frac{\theta }{2}\right)\cos \left(\frac{\theta }{2}\right)\mathrm{d}\theta$$$$=-2{\int }_0^{\frac{\pi }{2}}{\cos }^2\left(\frac{\theta }{2}\right)\mathrm{d}\theta =-{\int }_0^{\frac{\pi }{2}}(1+\cos \theta )\mathrm{d}\theta$$$$=-\frac{\pi }{2}-{\left[\sin \theta \right]}_0^{\frac{\pi }{2}}=-\frac{\pi }{2}-1$$$${\int }_0^1\sqrt{\frac{1-\mathrm{t}}{1+\mathrm{t}}}\mathrm{dt}{=}_{\mathrm{t}=\cos \theta }{\int }_{\frac{\pi }{2}}^0\sqrt{\frac{1-\cos \theta }{1+\cos \theta }}(-\sin \theta )\mathrm{d}\theta$$$$={\int }_0^{\frac{\pi }{2}}\sqrt{\frac{{2\sin }^2\left(\frac{\theta }{2}\right)}{{2\cos }^2\left(\frac{\theta }{2}\right)}}\sin \theta \mathrm{d}\theta$$$$={\int }_0^{\frac{\pi }{2}}\frac{\sin \left(\frac{\theta }{2}\right)}{\cos \left(\frac{\theta }{2}\right)}\times 2\sin \left(\frac{\theta }{2}\right)\cos \left(\frac{\theta }{2}\right)\mathrm{d}\theta$$$$=2{\int }_0^{\frac{\pi }{2}}{\sin }^2\left(\frac{\theta }{2}\right)\mathrm{d}\theta ={\int }_0^{\frac{\pi }{2}}(1-\cos \left(\theta \right))\mathrm{d}\theta$$$$=\frac{\pi }{2}-{\left[\sin \theta \right]}_0^{\frac{\pi }{2}}=\frac{\pi }{2}-1\Rightarrow$$$$\mathrm{I}=-\frac{\pi }{2}-1+\frac{\pi }{2}-1\Rightarrow \mathrm{I}=-2$$

Commented by SANOGO last updated on 27/Dec/21

$$thankyou$$

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