Evaulate $\int_0^1 x\sqrt{\frac{1-x^2}{1+x^2}}dx $ I have been challanged by my teacher to solve this integral,
However he gave me no hints, and I have no idea how to start 
$$\int_0^1 x\sqrt{\frac{1-x^2}{1+x^2}}dx $$
I noticed that putting $x=1-x$ only expands the term and doesnt really help, So anyone got any other idea,
I did try by Parts, It only makes it more messier 
 A: Substitute $x^2\to x$.  Then, 
$$\int_0^1x\sqrt{\frac{1-x^2}{1+x^2}}\,dx=\frac12 \int_0^1\sqrt{\frac{1-x}{1+x}}\,dx$$
Now, letting $x\to \cos x$ and using $1-\cos x=2\sin^2(x/2)$ and $1+\cos x=2\cos^2(x/2)$ yields
$$\begin{align}
\frac12 \int_0^1\sqrt{\frac{1-x}{1+x}}\,dx&=\frac12 \int_0^{\pi/2} \sin x\tan (x/2)\,dx\\\\
&=\int_0^{\pi/2} \sin^2 (x/2)\,dx\\\\
&=\frac12\int_0^{\pi/2}(1-\cos x)\,dx\\\\
&=\pi/4-1/2
\end{align}$$
A: HINT:
$$\arccos(x^2)=2y\implies0\le2y\le\pi, x=0\implies2y=\dfrac\pi2, x=1\implies2y=0$$
$$\implies\cos2y=x^2$$ 
and $$\sqrt{\dfrac{1-x^2}{1+x^2}}=+\tan y$$ as $0\le y\le\dfrac\pi4$
A: HINT : $$I=\int_{0}^{1}\frac{x-x^3}{\sqrt{1-x^4}}dx=\int_{0}^{1}\frac{xdx}{\sqrt{1-x^4}}+\frac{1}{4}\int_{0}^{1}\frac{-4x^3dx}{\sqrt{1-x^4}}$$
A: HINT:
$$\int_{0}^{1}x\sqrt{\frac{1-x^2}{1+x^2}}\space\text{d}x=$$

Substitute $u=x^2$ and $\text{d}u=2x\space\text{d}x$. New lower bound $u=0^2=0$ and upper bound $u=1^2=1$:

$$\frac{1}{2}\int_{0}^{1}\sqrt{\frac{1-u}{1+u}}\space\text{d}u=$$

Substitute $s=\frac{1-u}{1+u}$ and $\text{d}s=\left(-\frac{1-u}{(1+u)^2}-\frac{1}{1+u}\right)\space\text{d}u$. 
New lower bound $s=\frac{1-0}{1+0}=1$ and upper bound $s=\frac{1-1}{1+1}=0$:

$$-\int_{1}^{0}\frac{\sqrt{s}}{(-s-1)^2}\space\text{d}s=$$
$$\int_{0}^{1}\frac{\sqrt{s}}{(-s-1)^2}\space\text{d}s=$$

Substitute $p=\sqrt{s}$ and $\text{d}p=\frac{1}{2\sqrt{s}}\space\text{d}s$. 
New lower bound $p=\sqrt{0}=0$ and upper bound $p=\sqrt{1}=1$:

$$2\int_{0}^{1}\frac{p^2}{(-p^2-1)^2}\space\text{d}p=$$
$$2\int_{0}^{1}\frac{p^2}{(p^2+1)^2}\space\text{d}p$$
