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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.

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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 $u=x^{2}$.

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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.

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

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

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  • $\begingroup$ Thanks, this is great! Is it a part of a bigger idea or does it just happen to work here? $\endgroup$ – Bartek Nov 27 '12 at 20:11
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    $\begingroup$ Not really part of a bigger idea, except perhaps that roots are trouble and it is nice to get rid of them. And (in principle at least) every rational function of $\sin$ and $\cos$ can be integrated. It is very hard to dfine what one may mean by simplest approach. This is because one can "guess" the answer and then trivially verify it, or disguise the verification as a magic substitution. $\endgroup$ – André Nicolas Nov 27 '12 at 20:18
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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

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

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  • $\begingroup$ Very nice, thank you! $\endgroup$ – Bartek Nov 28 '12 at 9:34

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