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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?

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


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.

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the answer is supposed to be $\arctan(\tan^2x)+c$, which you can obtain by substituting $t=\tan^2x$

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  • $\begingroup$ His answer is correct, yours too, since $$\frac{\tan^2x+\cos2x}{1-\tan^2x\cos2x}=1$$ $\endgroup$ – Math-fun Jun 17 '15 at 12:49
  • $\begingroup$ @Math-fun:I like that! $\endgroup$ – David Quinn Jun 17 '15 at 12:52
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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, $$ \arctan\left(\tan^2(x)\right)+\arctan\left(\cos(2x)\right)=\frac\pi4 $$

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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.

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