<s>[INST] Find error in integration of $\int \frac {\sin 2x}{\sin^4 x + \cos^4 x} \, dx$? Find error in integration of $\int \frac {\sin 2x}{\sin^4 x + \cos^4 x}dx$?
The answer is supposed to be ($\arctan \tan^2 x + C$), but I obtained ($-\arctan \cos2x + C$) as follows. Please identify the error.
$$\int \frac {\sin 2x}{\sin^4 x + \cos^4 x}dx$$
$$= \int \frac {\sin 2x}{(\sin^2 x+ \cos^ 2x)^2 - 2\cos^2 x\sin^2 x}dx$$
$$= \int \frac {\sin 2x}{1- \frac{\sin^2 2x}{2}}dx$$
$$= \int \frac {2\sin 2x}{2- \sin^2 2x}dx$$
$$= \int \frac {2\sin 2x}{1 + \cos^2 2x}dx$$
$1 + \cos^2 2x = t ; dt = 2(\cos 2x)(-\sin 2x)(2)dx; 2\sin 2x.dx= \frac{-dt}{2\cos 2x} = \frac{-dt}{2(\sqrt{t-1})};$
$$\int \frac{\frac{-dt}{2\sqrt {t-1}}}{t} = -1/2\int \frac{dt}{t.\sqrt{t-1}}$$
$\sqrt{t-1} = u; du=\frac{dt}{2\sqrt{t-1}}; 2u.du = dt$;
$$= -1\int \frac{du}{u^2+1} = -\arctan{u} = -\arctan \sqrt{t-1} = -\arctan \sqrt{1+\cos^2 2x-1} = -\arctan \cos2x$$
Also, is there an easier method to this problem?
 [/INST] the answer is supposed to be $\arctan(\tan^2x)+c$, which you can obtain by substituting $t=\tan^2x$
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