Indefinite Integration of $x\sqrt{\frac{1-x}{1+x}}$

Question

I was wondering how to find out $$\int x\sqrt{\frac{1-x}{1+x}}\,dx$$

I was thinking of substitution $x=\sin^2t$, which helps with $1-x$. But then I cannot get rid of the denominator $1+x$.

Answer 1

HINT: Try the substitution

$$\frac{1-x}{1+x}=u^2$$

As Claude Leibovici hinted, that will remove the square root sign and give you a rational expression to integrate.

Answer 2

HINT... Try substituting $x=\cos 2\theta$ and then some fairly straightforward trig manipulation...

Answer 3

Let $x=\sin\theta, \;dx=\cos\theta\; d\theta$ to get

$\displaystyle\int x\sqrt{\frac{1-x}{1+x}}dx=\int\sin\theta\sqrt{\frac{1-\sin\theta}{1+\sin\theta}}\cos\theta d\theta=\int\sin\theta\sqrt{\frac{(1-\sin\theta)^2}{\cos^2\theta}}\cos\theta d\theta$

$\displaystyle=\int\sin\theta(1-\sin\theta) d\theta=\int\left(\sin\theta-\frac{1}{2}(1-\cos 2\theta)\right) d\theta=-\cos\theta-\frac{1}{2}\theta+\frac{1}{4}\sin2\theta+C$

$\displaystyle=-\cos\theta-\frac{1}{2}\theta+\frac{1}{2}\sin\theta\cos\theta+C=-\sqrt{1-x^2}-\frac{1}{2}\sin^{-1}x+\frac{1}{2}x\sqrt{1-x^2}+C$