Help with the integral $\int x\sqrt{\frac{1-x^2}{1+x^2}}dx$

Question

I would like to know what is $$\int x\sqrt{\frac{1-x^2}{1+x^2}}dx.$$ I put $x=\tan(y)$ to get integral of $\displaystyle \int \frac{\sin(y)}{\cos^3(y)}.\sqrt{\cos(2y)}dy$ I don't know whether $\sin(x)=t$ is a good substitution.

Answer 1

Hint. By the change of variable, $u=x^2$, $du=2xdx$, we have $$ \int x \sqrt{\frac{1-x^2}{1+x^2}}dx=\frac12\int \sqrt{\frac{1-u}{1+u}}du $$ then, by the change of variable, $$v=\sqrt{\frac{1-u}{1+u}}$$ we get $$ \int x \sqrt{\frac{1-x^2}{1+x^2}}dx=-2\int \frac{v^2}{(1+v^2)^2}dv $$ which is a rational function easier to evaluate.

Answer 2

Let $$I =\frac{1}{2} \int \sqrt{\frac{1-x^2}{1+x^2}}\cdot 2xdx$$

Now put $x^2=t\;,$ Then $2xdx = dt$

So $$I = \frac{1}{2}\int\sqrt{\frac{1-t}{1+t}}dt = \frac{1}{2}\int\sqrt{\frac{1-t}{1+t}\times \frac{1-t}{1-t}}dt$$

So $$I = \frac{1}{2}\int\frac{1}{\sqrt{1-t^2}}dt-\frac{1}{2}\int\frac{t}{\sqrt{1-t^2}}dt$$

So $$I = \frac{1}{2}\arcsin (t)+\frac{1}{2}\sqrt{1-t^2}+\mathcal{C}$$

So $$I = \frac{1}{2}\arcsin (x^2)+\frac{1}{2}\sqrt{1-x^4}+\mathcal{C}$$

Answer 3

In the same spirit as previous answer considering $$I = \int x\, \sqrt{\frac{1-x^2}{1+x^2}}\, dx$$ What I prefer is to get rid of the radical as early as possible.

So, let $$\sqrt{\frac{1-x^2}{1+x^2}}=u \implies x=\frac{\sqrt{1-u^2}}{\sqrt{u^2+1}}\implies dx=-\frac{2 u}{\sqrt{1-u^2} \left(u^2+1\right)^{3/2}}$$ This makes $$I=-\int\frac{2 u^2}{\left(u^2+1\right)^2}\,du$$ Now, $$\frac{ u^2}{\left(u^2+1\right)^2}=\frac{ u^2+1-1}{\left(u^2+1\right)^2}=\frac{1}{u^2+1}-\frac{1}{\left(u^2+1\right)^2}$$

Answer 4

HINT:

Metheod$\#1:$

Let $\sqrt{\dfrac{1-x^2}{1+x^2}}=y\implies \dfrac{1-x^2}{1+x^2}=y^2\implies x^2=\dfrac{1-y^2}{1+y^2}=\dfrac{2-(1+y^2)}{1+y^2}=\cdots$

$$2x\ dx=-2\dfrac{2y\ dy}{(1+y^2)^2}$$

$$\dfrac{d(y/1+y^2)}{dy}=?$$

Metheod$\#2:$

Let $x^2=\cos2y\implies0\le2y\le\pi$ and $\sqrt{\dfrac{1-x^2}{1+x^2}}=+\tan y$

$x\ dx=-2\sin2y\ dy=-4\sin y\cos y\ dy$