1
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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

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

$\endgroup$
8
$\begingroup$

$$\frac{1+\cos u}{1-\cos u}=2\cot^2\left(\frac{u}{2}\right).$$ 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.

$\endgroup$
7
$\begingroup$

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 $u=x^2\implies du=2xdx$

$\endgroup$
3
$\begingroup$

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

$\endgroup$
  • $\begingroup$ i said using $x = cos(u)$ $\endgroup$ – KEVIN DLL Sep 30 '18 at 18:46
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.