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Find the value of $$ \int_0^{2\pi}\frac{\mathrm dx}{1+\tan^3x} $$

My attempt: I split $\tan x$ as $\frac{\sin x}{\cos x}$, and got the following form $$ \int_0^{2\pi}\frac{\cos^3x\,\mathrm dx}{\sin^3x+\cos^3x} $$ and am stuck there.

Any help will be appreciated.

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  • $\begingroup$ Do you know how to use partial fractions to solve an integral? $\endgroup$ – Franklin Pezzuti Dyer Jul 11 '17 at 13:23
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    $\begingroup$ I wonder if $tan \text x$ is continuous on $\frac{\pi}{2}$ $\endgroup$ – Crazy Jul 11 '17 at 13:23
  • $\begingroup$ @Crazy It's in the denominator, so the entire fraction tends to $0$, no problem there. $\endgroup$ – Simply Beautiful Art Jul 11 '17 at 13:32
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    $\begingroup$ just a side note: the improper integral can be simplified to $$2\int_{y\pi}^{(y+1)\pi}\frac{\mathrm dx}{1+\tan^3 x}$$ for any $y\in\Bbb R$ because $\tan x$ is $\pi$-periodic and also it is the integrand. $\endgroup$ – Masacroso Jul 11 '17 at 13:39
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    $\begingroup$ This integral is divergent because at $3\pi/4$ and $7\pi/4$ it goes to $\infty$ and the integral function $F(x)=\frac{x}{2}+\frac{1}{6} \log (\tan (x)+1)+\frac{1}{4} \log \left(\tan ^2(x)+1\right)-\frac{1}{3}\log\left(\tan ^2(x)-\tan (x)+1\right)$ diverges $\endgroup$ – Raffaele Jul 11 '17 at 14:31
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PARTIAL SOLUTION: $$\int \frac{dx}{1+\tan^3 x}$$ $$=\int \frac{dx}{(1+\tan x)(1-\tan x+\tan^2 x)}$$ Then use the substitution $x \to \arctan u$: $$=\int \frac{du}{(1+u)(1-u+u^2)(1+u^2)}$$ Then proceed to use partial fractions. Can you take it from here?

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    $\begingroup$ Substitutions should be given by differentiable maps with a differentiable inverse, but $\tan(x)$ is not injective over $(0,2\pi)$. $\endgroup$ – Jack D'Aurizio Jul 12 '17 at 10:26
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HINT: the partial decomposition of your integral is given by $$1/3\,{\frac {-2\,u+1}{{u}^{2}-u+1}}+1/6\, \left( 1+u \right) ^{-1}+1/2 \,{\frac {1+u}{{u}^{2}+1}} $$ but the definite integral doesn't converge on the given interval

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  • $\begingroup$ The partial fraction decomposition (of what?) is irrelevant for proving the integral is divergent. $\endgroup$ – Jack D'Aurizio Jul 12 '17 at 10:26
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The integrand function has simple poles (non-integrable singularities) at $x\in\left\{\frac{3\pi}{4},\frac{7\pi}{4}\right\}\subset\left(0,2\pi\right)$ hence the integral is not convergent in the usual sense.
Its principal value, however, is finite:

$$\begin{eqnarray*}\text{PV}\int_{0}^{2\pi}\frac{dx}{1+\tan^3(x)} &=& \int_{0}^{3\pi/4}\left(\frac{1}{1+\tan^3\left(\frac{3\pi}{4}+x\right)}+\frac{1}{1+\tan^3\left(\frac{3\pi}{4}-x\right)}\right)\,dx\\&+&\int_{0}^{\pi/4}\left(\frac{1}{1+\tan^3\left(\frac{7\pi}{4}+x\right)}+\frac{1}{1+\tan^3\left(\frac{7\pi}{4}-x\right)}\right)\,dx \end{eqnarray*}$$ greatly simplifies into: $$ \text{PV}\int_{0}^{2\pi}\frac{dx}{1+\tan^3(x)} = \int_{0}^{3\pi/4}1\,dx+\int_{0}^{\pi/4}1\,dx = \color{red}{\pi}.$$

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