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?
 A: $$\int\frac{\sin(2x)}{\sin^4(x)+\cos^4(x)}dx=\int\frac{4\sin(2x)}{\left(1-\cos(2x)\right)^2+(1+\cos^2(2x))^2}dx=-\int\frac{\,d\cos(2x)}{1+\cos^2(2x)}\\
=-\arctan(\cos(2x))+C.$$

As
$$\tan^2(x)=\frac{\sin^2(x)}{\cos^2(x)}=\frac{1-\cos(2x)}{1+\cos(2x)}=\tan\left(\frac\pi4-\arctan(\cos(2x))\right),$$
$\arctan(\tan^2(x))$ and $-\arctan(\cos(2x))$ only differ by a constant.
A: the answer is supposed to be $\arctan(\tan^2x)+c$, which you can obtain by substituting $t=\tan^2x$
A: Since
$$
\begin{align}
1
&=(\sin^2(x)+\cos^2(x))^2\\[6pt]
&=\sin^4(x)+\cos^4(x)+2\sin^2(x)\cos^2(x)\\[2pt]
&=\sin^4(x)+\cos^4(x)+\frac12\sin^2(2x)
\end{align}
$$
we can proceed with partial fractions
$$
\begin{align}
&\int\frac{\sin(2x)}{\sin^4(x)+\cos^4(x)}\,\mathrm{d}x\\[6pt]
&=\int\frac{\sin(2x)}{1-\frac12\sin^2(2x)}\,\mathrm{d}x\\
&=\frac1{\sqrt2}\int\left(\frac1{1-\frac1{\sqrt2}\sin(2x)}-\frac1{1+\frac1{\sqrt2}\sin(2x)}\right)\,\mathrm{d}x\\
&=\frac1{\sqrt2}\int\left(\frac1{1-\frac{\sqrt2\tan(x)}{1+\tan^2(x)}}-\frac1{1+\frac{\sqrt2\tan(x)}{1+\tan^2(x)}}\right)\frac{\mathrm{d}\tan(x)}{1+\tan^2(x)}\\
&=\frac1{\sqrt2}\int\left(\frac1{\tan^2(x)-\sqrt2\tan(x)+1}-\frac1{\tan^2(x)+\sqrt2\tan(x)+1}\right)\,\mathrm{d}\tan(x)\\
&=\int\left(\frac1{(\sqrt2\tan(x)-1)^2+1}-\frac1{(\sqrt2\tan(x)+1)^2+1}\right)\,\mathrm{d}\sqrt2\tan(x)\\[12pt]
&=\arctan(\sqrt2\tan(x)-1)-\arctan(\sqrt2\tan(x)+1)+C+\tfrac\pi2\\[9pt]
&=\arctan\left(\frac{-1}{\tan^2(x)}\right)+C+\tfrac\pi2\\[9pt]
&=\bbox[5px,border:2px solid #C0A000]{\arctan\left(\tan^2(x)\right)+C}
\end{align}
$$
We can also proceed by writing everything in terms of $\cos(2x)$
$$
\begin{align}
\int\frac{\sin(2x)}{\sin^4(x)+\cos^4(x)}\,\mathrm{d}x
&=\int\frac{\sin(2x)}{1-\frac12\sin^2(2x)}\,\mathrm{d}x\\
&=-\int\frac1{1+\cos^2(2x)}\,\mathrm{d}\cos(2x)\\[9pt]
&=\bbox[5px,border:2px solid #C0A000]{-\arctan(\cos(2x))+C}
\end{align}
$$

These two answers are copacetic since
$$
\begin{align}
\tan\left(\arctan\left(\tan^2(x)\right)+\arctan(\cos(2x))\right)
&=\frac{\tan^2(x)+\cos(2x)}{1-\tan^2(x)\cos(2x)}\\[6pt]
&=\frac{\sin^2(x)+\cos^2(x)(\cos^2(x)-\sin^2(x))}{\cos^2(x)-\sin^2(x)(\cos^2(x)-\sin^2(x))}\\[6pt]
&=\frac{\sin^2(x)+\cos^4(x)-\sin^2(x)\cos^2(x)}{\cos^2(x)+\sin^4(x)-\sin^2(x)\cos^2(x)}\\[6pt]
&=\frac{\sin^4(x)+\cos^4(x)}{\cos^4(x)+\sin^4(x)}\\[12pt]
&=1
\end{align}
$$
That is,
$$
\arctan\left(\tan^2(x)\right)+\arctan\left(\cos(2x)\right)=\frac\pi4
$$
A: You have the correct answer. See for example here: http://www.integral-calculator.com/#
This site will give answer $$\arctan(2\sin^2(x)-1)+C=\arctan(2\sin^2(x)-\sin^2(x)-\cos^2(x))+C=\arctan(-\cos(2x))+C=-\arctan(\cos(2x))+C$$
Wolfram Alpha agrees with you also.
