evaluation of $\int_{0}^{1}\frac{1-x^2}{\left(1+x^2\right)\sqrt{1+x^4}}dx$ $\displaystyle \int_{0}^{1}\frac{1-x^2}{\left(1+x^2\right)\sqrt{1+x^4}}dx$
By Using Substution $\sqrt{1+x^4} = (1+x^2)\cdot \cos \theta$
I have Tried it without using the given substution.
$\bf{My\; Try}::$ $\displaystyle \int_{0}^{1}\frac{1-x^2}{\left(1+x^2\right)\sqrt{1+x^4}}dx = \int_{0}^{1}\frac{1-x^2}{x\cdot \left(x+\frac{1}{x}\right)\cdot x \cdot \sqrt{\left(x^2+\frac{1}{x^2}\right)}}dx$
$\displaystyle = -\displaystyle \int_{0}^{1}\frac{\left(1-\frac{1}{x^2}\right)}{\left(x+\frac{1}{x}\right)\cdot \sqrt{\left(x+\frac{1}{x}\right)^2-\left(\sqrt{2}\right)^2}}dx$
Now Let $\displaystyle \left(x+\frac{1}{x}\right) = t$ and $\displaystyle \left(1-\frac{1}{x^2}\right)dx = dt$
$\displaystyle = -\int_{0}^{1}\frac{1}{t\sqrt{t^2-\left(\sqrt{2}\right)^2}}dt = -\frac{1}{\sqrt{2}}\left[\sec^{-1}\left(\frac{x+\frac{1}{x}}{\sqrt{2}}\right)\right]_{0}^{1} = -\frac{1}{\sqrt{2}}\left(\sec^{-1}\left(\sqrt{2}\right)-\sec^{-1}\left(\infty \right)\right)$
$\displaystyle = \frac{1}{\sqrt{2}}\left(\frac{\pi}{2}-\frac{\pi}{4}\right) = \frac{\pi}{4\sqrt{2}}$
Would anyone explain me how can i solve using the substution  $\sqrt{1+x^4} = (1+x^2)\cdot \cos \theta$
Help Required
Thanks

Nicely done. So they suggest setting 
$$
\cos \theta =\frac{\sqrt{1+x^4}}{1+x^2} \quad\Rightarrow\quad \frac{2x(1-x^2)}{\sqrt{1+x^4}(1+x^2)^2}dx=\sin \theta \,d\theta
$$
and, since $\sin\theta\geq 0$ here:
$$
\sin\theta =\sqrt{1-\cos^2\theta}=\frac{\sqrt{2} \;x}{1+x^2} \quad\Rightarrow \quad \frac{1-x^2}{(1+x^2)\sqrt{1+x^4}} dx=\frac{d\theta}{\sqrt{2}}
$$
This is indeed a valid change of variable $\theta =\arccos\left( \frac{\sqrt{1+x^4}}{1+x^2}\right)$. The bounds become $\arccos(1)=0$ and $\arccos(\sqrt{2}/2)=\pi/4$. This yields
$$
\int_0^1 \frac{1-x^2}{(1+x^2)\sqrt{1+x^4}} dx=\int_0^{\pi/4}\frac{d\theta}{\sqrt{2}}=\frac{\pi}{4\sqrt{2}}
$$
