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I have to integrate $\int \sqrt{\frac{1+x}{1-x}}$ using $x = \cos(u)\Rightarrow dx = -\sin(u) du$

My attempts: $$\int \sqrt{\frac{1+x}{1-x}}dx=\int \sqrt{\frac{(1+x)(1-x)}{(1-x)(1-x)}}dx=\int \frac{\sqrt{1-x^2}}{1-x}dx=\int \frac{\sqrt{1-x^2}}{1-x}dx$$ $$\int \frac{\sqrt{\sin^2(u)}}{1-\cos(u)}( -\sin(u))du=\int \frac{-\sin(u)^2}{1-\cos(u)}du=\int \frac{-(1 - \cos^2(u)) }{1-\cos(u)}du$$ , someone could help because i block here because i don't get the same resultat with this website

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5 Answers 5

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$$\frac{1+\cos u}{1-\cos u}=2\cot^2\left(\frac{u}{2}\right).$$ So your integral is $$\int\sqrt{\frac{1+x}{1-x}}\, dx=-\sqrt{2}\int \sin u \cot\left(\frac{u}{2}\right) \, du=-2\sqrt{2}\int \cos^2 \left(\frac{u}{2}\right)\, du$$ Here we are using $\sin u =2\sin \left(\frac{u}{2}\right)\cos \left(\frac{u}{2}\right)$. Now once again we can write $2\cos^2 \left(\frac{u}{2}\right)=1+\cos u$ to complete the integral.

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Try this-

$$\int\sqrt{\frac{1+x}{1-x}}dx=\int\frac{1+x}{\sqrt{1-x^2}}dx=\int\frac{1}{\sqrt{1-x^2}}+\frac{x}{\sqrt{1-x^2}}dx$$

The first integral is $\arcsin x$ and the second is evaluated by substituting $u=x^2\implies du=2xdx$

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There are many ways, and a natural temptation is to exploit the fact that $g(z)=\frac{1-z}{1+z}$ is an involution, $g(g(z))=z$. In particular $$ \int\sqrt{\frac{1+x}{1-x}}\,dx = \int \frac{dx}{\sqrt{g(x)}}\stackrel{x\mapsto g(z)}{=}\int\frac{g'(z)\,dz}{\sqrt{z}}\stackrel{z\mapsto u^2}{=}2\int g'(u^2)\,du =4\int\frac{du}{(1+u^2)^2}$$ and the last integral is elementary. By partial fraction decomposition or IBP $$ \int\frac{du}{(1+u^2)^2}=C+\frac{1}{2}\left(\frac{u}{1+u^2}+\arctan u\right) $$ and now it is enough to substitute $u=\sqrt{g(x)}$ in the RHS.

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$$\int \frac{\sqrt{\sin^2(u)}}{1-\cos(u)}( -\sin(u))du=\int \frac{-\sin^2(u)}{1-\cos(u)}du=\int \frac{-(1 - \cos^2(u)) }{1-\cos(u)}du$$ $$\int \frac{-(1 - \cos(u))(1 + \cos(u)) }{1-\cos(u)}du=-\int (1 + \cos(u)) du=-u-\sin u+C$$

Can you replace the value of $u$ by using $\cos u =x$

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Substitute $$t=\sqrt{\frac{1+x}{1-x}}$$ then we get $$x=\frac{t^2-1}{t^2+1}$$ $$dx=\frac{4t}{(t^2+1)^2}dt$$

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  • $\begingroup$ i said using $x = cos(u)$ $\endgroup$
    – KEVIN DLL
    Sep 30, 2018 at 18:46

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