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Question Number 124632 by liberty last updated on 05/Dec/20

$$Calculate\underset{0}{\stackrel{2}{\int }}\sqrt{\frac{2+x}{2-x}}dx$$

Answered by bemath last updated on 05/Dec/20

$$putx=2\cos 2\theta$$$$I=\underset{\pi /4}{\stackrel{0}{\int }}\sqrt{\frac{2+2\cos 2\theta }{2-2\cos 2\theta }}(-4\sin 2\theta )d\theta$$$$I=4\underset{0}{\stackrel{\pi /4}{\int }}\sqrt{\frac{1+\cos 2\theta }{1-\cos 2\theta }}\left(2\sin \theta \cos \theta \right)d\theta$$$$I=4\underset{0}{\stackrel{\pi /4}{\int }}2\cos {}^2\theta d\theta$$$$I=4{[\frac{\sin 2\theta }{2}+\theta ]}_0^{\pi /4}=4(\frac{1}{2}+\frac{\pi }{4})$$$$I=2+\pi$$

Answered by mathmax by abdo last updated on 05/Dec/20

$$\mathrm{A}={\int }_0^2\sqrt{\frac{2+\mathrm{x}}{2-\mathrm{x}}}\mathrm{dx}\mathrm{we}\mathrm{do}\mathrm{the}\mathrm{changement}\mathrm{x}=2\mathrm{cost}\Rightarrow$$$$\mathrm{A}={\int }_{\frac{\pi }{2}}^0\sqrt{\frac{1+\mathrm{cost}}{1-\mathrm{cost}}}(-2\mathrm{sint})\mathrm{dt}=2{\int }_0^{\frac{\pi }{2}}\frac{\cos \left(\frac{\mathrm{t}}{2}\right)}{\sin \left(\frac{\mathrm{t}}{2}\right)}\times 2\cos \left(\frac{\mathrm{t}}{2}\right)\sin \left(\frac{\mathrm{t}}{2}\right)\mathrm{dt}$$$$=4{\int }_0^{\frac{\pi }{2}}{\cos }^2\left(\frac{\mathrm{t}}{2}\right)\mathrm{dt}=4{\int }_0^{\frac{\pi }{2}}\frac{1+\mathrm{cost}}{2}\mathrm{dt}=2{\int }_0^{\frac{\pi }{2}}(1+\mathrm{cost})\mathrm{dt}$$$$=\pi +2{\left[\mathrm{sint}\right]}_0^{\frac{\pi }{2}}=\pi +2$$

Answered by MJS_new last updated on 05/Dec/20

$$\underset{0}{\stackrel{2}{\int }}\sqrt{\frac{2+x}{2-x}}dx=$$$$[t=\sqrt{\frac{2-x}{2+x}}\to dx=-\frac{1}{2}{(x+2)}^{3/2}\sqrt{2-x}dt]$$$$=8\underset{0}{\stackrel{1}{\int }}\frac{dt}{{(t^2+1)}^2}={[\frac{4t}{t^2+1}+4\arctan t]}_0^1=\pi +2$$

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