Integrate $ \int \frac{1}{\sin^{4}x+\cos^{4}x}dx $ Show that$$ \int \frac{1}{\sin^{4}(x)+\cos^{4}(x)}dx \ = \frac{1}{\sqrt{2}}\arctan\left(\frac{\tan2x}{\sqrt{2}}\right)+C$$
I have tried using Weierstrass substitution but I can't seem to get to the answer... Should I be using the said method or is there another way I can approach the question? Since the integrand evaluates into an arctangent function I am assuming there is some trickery in the manipulation that can get me there. But I just can't seem to see it...

Original post was to evaluate $$\int\frac{\,dx}{\sin^4(x)\cos^4(x)}$$ but I believe that the intended integral was $$\int\frac{\,dx}{\sin^4(x)+\cos^4(x)}$$ which I evaluate after the original request.
$$\begin{align}I=&\int\frac{\,dx}{\sin^4(x)\cos^4(x)}\\&=2^4\int\frac{\,dx}{\bigr[2\sin(x)\cos(x)\bigr]^4}\\&=2^4\int\frac{\,dx}{\sin^4(2x)}\\&=2^4\int\csc^4(2x)\,dx\\&=2^3\int \csc^4(u)\,du\end{align}$$
Using the reduction formula for cosecant, $$\int\csc^{m}(x)\,dx=-\frac{\csc^{m-1}(x)\cos(x)}{m-1}+\frac{m-2}{m-1}\int\csc^{m-2}(x)\,dx$$ or more concisely, $$J_m= -\frac{\csc^{m-1}(x)\cos(x)}{m-1}+\frac{m-2}{m-1}J_{m-2} $$ then we can obtain $$J_4=-\frac{\csc^{3}(u)\cos(u)}{3}+\frac{2}{3}J_2$$ where $J_2=-\cot(u)+C$
Let $$J_4=\int\csc^4{u}\,du$$ so that $$\begin{align}I&=2^3J_4\\&=2^3\bigg[-\frac{\csc^{3}(u)\cos(u)}{3}-\frac{2}{3}\cot(u)\bigg] +C\\&=-\frac{2^3}{3}\bigg[\csc^{3}(2x)\cos(2x)+2\cot(2x)\bigg] +C\\&=-\frac{2^3}{3}\cot(2x)\bigg[\csc^2(2x)+2\bigg]+C\end{align}$$
If you don't know the reduction formula off-hand or don't feel like deriving it, expand the integrand as $$\csc^4(u)=\csc^2(u)\csc^2(u)=\csc^2(u)(1+\cot^2(u))=\csc^2(u)+\csc^2(u)\cot^2(u)$$ which is fairly simple to integrate.
Now to evaluate $$I=\int\frac{\,dx}{\sin^4(x)+\cos^4(x)}$$
$$\begin{align}I&=\int\frac{\,dx}{\cos^4(x)-2\sin^2(x)\cos^2(x)+\sin^4(x)+2\sin^2(x)\cos^2(x)}\\&=\int\frac{\,dx}{[\cos^2(x)-\sin^2(x)]^2+2\sin^2(x)\cos^2(x)}\\&=\int\frac{\,dx}{\cos^2(2x)+\frac{\sin^2(2x)}{2}}\\&=\int\frac{\sec^2(2x)}{1+\frac{\tan^2(2x)}{2}}\,dx\\&=\int\frac{\sec^2(2x)}{1+\bigg(\frac{\tan(2x)}{\sqrt{2}}\bigg)^2}\,dx \end{align} $$
Let $u=\frac{\tan(2x)}{\sqrt{2}}$ and $\,du=\frac{2}{\sqrt{2}}\sec^2(2x)\,dx$. Then $$\begin{align}I&=\frac{\sqrt{2}}{2}\int\frac{\,du}{1+u^2}\\&=\frac{1}{\sqrt{2}}\arctan(u)+C\\&=\frac{1}{\sqrt{2}}\arctan\bigg(\frac{\tan(2x)}{\sqrt{2}}\bigg)+C \end{align} $$
