$$\displaystyle \int \frac{1}{\sin^4 x + \cos^4 x}dx$$
My attempt:
$\displaystyle \int \frac{1}{1-2\sin^2 x \cos^2 x}dx$
$\displaystyle \int \frac{1}{1-\frac{\sin^2 2x}{2}}dx$
$2\displaystyle \int \frac{1}{2\cos^2 2x + \sin^2 2x}dx$
$2\displaystyle \int \frac{\sec^2 2x}{2+\tan^2 2x}dx$
$t = \tan 2x$
Any help would be appreciated. $\displaystyle \frac{dt}{dx} = (\sec^2 2x)2$
$\displaystyle \int \frac{1}{2+t^2}dt$
$\displaystyle \frac{1}{\sqrt2}\tan^{-1}\bigg(\frac{t}{\sqrt2}\bigg)$ + C
$\displaystyle \frac{1}{\sqrt2}\tan^{-1}\bigg(\frac{\tan2x}{\sqrt2}\bigg)$ + C
But the answer given in my book is,
$\displaystyle \frac{1}{\sqrt2}\tan^{-1}\bigg(\frac{\sqrt2}{\tan 2x}\bigg)$+ C
Any help would be appreciated.
