What is the easiest way to integrate $\left(\frac{1-x}{1+x}\right)^{1/2}?$

Question

This is an indefinite integral that's supposed to be very easy:

$$I=\int\sqrt{\frac{1-x}{1+x}}\,dx$$

I can only think of one way of calculating it, and it's a bit complicated, that is:

• substitute $x=\sin u$, and obtain $dx=(\cos u)\,du$ and $$I=\int\sqrt{\frac{1-\sin u}{1+\sin u}}(\cos u) \,du=\int\frac{1-\tan\frac{u}{2}}{1+\tan\frac{u}{2}}(\cos u)\,du.$$

• substitute $t=\tan \dfrac u2$, obtaining $du=\dfrac{2\,dt}{1+t^2},$ $\cos u=\dfrac{1-t^2}{1+t^2}$ and $$I=\int \frac{1-t}{1+t}\frac{1-t^2}{1+t^2}\frac2{1+t^2}\,dt=2\int\left(\frac{1-t}{1+t}\right)^2\,dt,$$

which can be calculated. I'm not even sure if this is correct though. But even if it is, I think this way is too difficult for the place in which I found this integral, which is a set of indefinite integrals where obvious substitutions work and no knowledge is necessary beyond how substitution works in general. I think there must be an easy way to do it that I don't see.

Answer 1

Multiply the top and bottom of your square root's argument by $1-x$ to get: $$I=\int\sqrt{\frac{(1-x)^{2}}{1-x^{2}}}dx=\int\frac{1-x}{\sqrt{1-x^{2}}}dx=\int\frac{1}{\sqrt{1-x^{2}}}dx-\int\frac{x}{\sqrt{1-x^{2}}}dx$$ The first integral yields to the subsitution $x=\sin(\theta)$ and the second to $u=x^{2}$.

Answer 2

How about this: $$ \int \sqrt{\frac{1-x}{1+x}} dx = \int \sqrt{ \frac{(1-x)^2}{1-x^2}} dx = \int \frac{1-x}{\sqrt{1-x^2}} dx$$ which is solvable by u-subst. and inverse sine.

Answer 3

Substitute $u = \sqrt{(1-x)/(1+x)}$, and you get a rational integrand.

Answer 4

Use the identity $\cos 2\theta=2\cos^2\theta-1=1-2\sin^2\theta$. Let $x=\cos 2\theta$.

Answer 5

There was an alternate way you could have proceeded.

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

Also, if you know your trig identities, making the substitution $x=\cos u$ yields

$$\int-\sqrt{\frac{1-\cos u}{1+\cos u}}\sin udu=-\int\tan\frac u2\sin udu=$$ $$-\int\left(\frac{1-\cos u}{\sin u}\right)\sin udu$$