<s>[INST] 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...
 [/INST] $$\sin ^4(x)+\cos ^4(x)=\left(\sin ^2(x)+\cos ^2(x)\right)^2-2\sin^2x\cos^2x=1-\frac{1}{2}\sin^2(2x)=\\=1-\frac{1}{2}\cdot\frac{1-\cos(4x)}{2}=\frac{1}{4}\left(3+\cos(4x)\right)$$
$$\cos(4x)=\frac{1-\tan^2(2x)}{1+\tan^2(2x)}=\frac{1-t^2}{1+t^2}$$
where $\tan(2x)=t$, $x=\frac{1}{2}\arctan t$, $dx=\frac{dt}{2(1+t^2)}$
$$\int \frac{1}{\sin ^4(x)+\cos ^4(x)} \, dx=\int \frac{4}{3+\cos(4x)} \, dx=$$
$$=4\int \frac{1}{3+4\frac{1-t^2}{1+t^2}}\cdot \frac{dt}{2(1+t^2)}=\int \frac{dt}{2+t^2}$$
$t=u\sqrt{2}$, $dt=du\sqrt 2$
$$\int \frac{dt}{2+t^2}=\int \frac{du\sqrt 2}{2+2u^2}=\frac{\sqrt 2}{2}\int\frac{du}{1+u^2} = \frac{\sqrt 2}{2}\arctan u+C=\frac{\sqrt 2}{2}\arctan \frac{t}{\sqrt 2}+C=$$
$$=\frac{\sqrt 2}{2}\arctan \frac{\tan(2x)}{\sqrt 2}+C$$
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