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The following problem came up in my last examination.

$$ \int {\frac{1}{1+ \tan^4 x} dx}$$

The difficulty I was facing was that I wasn't able to find anything to substitute as there was nothing special in the numerator. So I tried the following approach:

$$ \int {\frac{1}{1+ \tan^4 x}}dx = \int {\frac{\cos^4 x}{\cos^4 x + \sin^4 x}}dx$$

However this came to no good as it was still useless to substitute $\sin x$ as $t$ and get $dt=\cos x dx$

Next I tried substituting $\tan x$ as $t$
$$ t=\tan x \implies dt=\sec^2 x dx \implies dx=\frac{dt}{1+t^2} $$

giving me

$$ \int {\frac{dx}{(1+t^4)(1+t^2)}}$$

My question then is how do I split these into partial fractions and solve the integral? And, of course, if there is a better way to solve this integral, please suggest one.

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  • $\begingroup$ Would $u=\tan^4(x)$ help? $\endgroup$ – Permian Jan 17 '15 at 18:21
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Duplication formulas and the relation $\frac{1}{2-u^2}=\frac{1}{2\sqrt{2}}\left(\frac{1}{\sqrt{2}-u}-\frac{1}{\sqrt{2}+u}\right)$ give a good way:

$$ I = \int\frac{\cos^4 t}{\sin^4 t+\cos^4 t}\,dt=\int\frac{\cos^4 t}{1-2\sin^2t\cos^2 t}=\int\frac{\left(\frac{1+\cos(2t)}{2}\right)^2}{1-\frac{1}{2}\sin^2(2t)}\,dt$$ hence: $$ I = \frac{1}{2}\int\frac{(1+\cos(2t))^2}{1+\cos^2(2t)}\,dt=\frac{t}{2}+\int\frac{\cos(2t)}{2-\sin^2(2t)}\,dt$$ and: $$ I = \color{red}{\frac{t}{2}+\frac{1}{4\sqrt{2}}\log\frac{\sqrt{2}+\sin(2t)}{\sqrt{2}-\sin(2t)}}.$$

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    $\begingroup$ Nicely done! It's a way of going that does not feel so intuitive for me, but it surely was effective here. $\endgroup$ – mickep Jan 17 '15 at 18:57
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    $\begingroup$ Nice answer -- this is easier than using partial fractions. $\endgroup$ – user84413 Jan 17 '15 at 19:11
  • $\begingroup$ Nice Answer! Thanks for the help and yes it is easier than partial fractions. $\endgroup$ – Ris97 Jan 18 '15 at 6:42
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HINT:

$$\frac2{(1+t^2)(1+t^4)}=\frac{1-t^4+1+t^4}{(1+t^2)(1+t^4)}=\frac{1-t^2}{1+t^4}+\frac1{1+t^2}$$

For $\dfrac{1-t^2}{1+t^4}=\dfrac{1-1/t^2}{t^2+1/t^2},$

as $\int(1-1/t^2)dt=t+1/t,$ write $t^2+1/t^2=(t+1/t)^2-2$

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  • $\begingroup$ Wow! why didn't I think of that! $\endgroup$ – Ris97 Jan 18 '15 at 6:43
  • $\begingroup$ @Ris97, Could you follow the method? $\endgroup$ – lab bhattacharjee Jan 18 '15 at 7:57
  • $\begingroup$ yes I have understood your method. $\endgroup$ – Ris97 Jan 23 '15 at 18:40
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hint we have $$1+t^4=(t^2)^2+2t^2+1-2t^2=(t^2+1)^2-2t^2=(t^2+1-\sqrt{2}t(t^2+1+\sqrt{2}t)$$ then you must prove that $$\frac{1}{(1+t^4)(1+t^2)}=1/4\,{\frac {t\sqrt {2}-1}{t\sqrt {2}-{t}^{2}-1}}+1/2\, \left( {t}^{2} +1 \right) ^{-1}+1/4\,{\frac {t\sqrt {2}+1}{t\sqrt {2}+{t}^{2}+1}} $$

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    $\begingroup$ What is the next step? $\endgroup$ – lab bhattacharjee Jan 17 '15 at 18:37

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