Question

n%28x%29%20-%20%20%5Ctan%20%7B%7D%5E%7B2%7D%20%28x%29%20%20%7D%20%20%20%5C%3A%20dx%20%5C%5C%20" id="TexFormula1" title=" \rm \int_{ 0}^ {\large\frac{\pi}4} \sqrt{ \tan(x) - \tan {}^{2} (x) } \: dx \\ " alt=" \rm \int_{ 0}^ {\large\frac{\pi}4} \sqrt{ \tan(x) - \tan {}^{2} (x) } \: dx \\ " align="absmiddle" class="latex-formula">

Answer 1

First substitute $x=\tan^{-1}(y)$ to rewrite the integral as

$\displaystyle \int_0^{\pi/4} \sqrt{\tan(x) - \tan^2(x)} \, dx = \int_0^1 \frac{\sqrt{y-y^2}}{1+y^2} \, dy$

Now use an Euler substitution, $z=\frac{\sqrt{y-y^2}}y$ to rewrite it again as

$\displaystyle \int_0^{\pi/4} \sqrt{\tan(x) - \tan^2(x)} \, dx = 2 \int_0^\infty \frac{t^2}{(t^2+1)^2 + 1) (t^2 + 1)} \, dt$

where we take

$\sqrt{y - y^2} = \sqrt{-y(y-1)} = yt \implies y = \dfrac1{1+t^2} \text{ and } dy = -\dfrac{2t}{(1+t^2)^2} \, dt$

Partial fractions:

$\displaystyle \frac{t^2}{((t^2+1)^2+1) (t^2 + 1)} = \dfrac{t^2+2}{t^4+2t^2+2} - \dfrac1{t^2+1}$

so that

$\displaystyle \int_0^{\pi/4} \sqrt{\tan(x) - \tan^2(x)} \, dx = 2 \left(\int_0^\infty \frac{t^2+2}{t^4+2t^2+2} \, dt - \int_0^\infty \frac{dt}{t^2+1}\right)$

The second integral is trivial,

$\displaystyle \int_0^\infty \frac{dt}{t^2+1} = \lim_{t\to\infty}\tan^{-1}(t) - \tan^{-1}(0) = \frac\pi2$

For the other, I'm compelled to use the residue theorem, though real methods are doable too (e.g. trig substitution). Consider the contour integral

$\displaystyle \int_\Gamma f(z) \, dz = \int_\Gamma \frac{z^2+2}{z^4+2z^2+2} \, dz$

where $\Gamma$ is a semicircle in the upper half of the complex plane, and its diameter lies on the real axis connecting $-R$ to $R$. The value of this integral is 2πi times the sum of the residues in the upper half-plane. It's fairly straightforward to convince ourselves that the integral along the circular arc vanishes as $R\to\infty$, so the contour integral converges to the integral over the entire real line. Note that

$\displaystyle 2 \int_0^\infty \frac{t^2+2}{t^4+2t^2+2} \, dt = \int_{-\infty}^\infty \frac{t^2+2}{t^4+2t^2+2} \, dt$

since the integrand is even.

Find the poles of $f(z)$.

$z^4 + 2z^2 + 2 = 0 \\\\ ~~~~ \implies (z^2+1)^2 = -1 \\\\ ~~~~ \implies z^2 = -1 \pm i \\\\ ~~~~ \implies z = \pm \sqrt{-1 \pm i} = \sqrt[4]{2}\, e^{\pm i(3\pi/8 + \pi k)}$

where $k\in\{0,1\}$.

The two poles we care about are at $z_1=\sqrt[4]{2}\,e^{i\,3\pi/8}$ and $z_2=\sqrt[4]{2}\,e^{-i\,11\pi/8}$. Compute the residues at each one.

$\displaystyle \mathrm{Res}\left\{f(z),z=z_1\right\} = \lim_{z\to z_1} \frac{f(z)}{z-z_1} = -\frac1{2^{7/4}}\,ie^{-i\,\pi/8} \\\\ ~~~~~~~~~~~~~~~~~~~~~~~~= -\frac1{2^{7/4}} \left(\sin\left(\frac\pi8\right) + i \cos\left(\frac\pi8\right)\right)$

$\displaystyle \mathrm{Res}\left\{f(z),z=z_2\right\} = \lim_{z\to z_2} \frac{f(z)}{z-z_2} = -\frac1{2^{7/4}}\,ie^{i\,\pi/8} \\\\ ~~~~~~~~~~~~~~~~~~~~~~~~= \frac1{2^{7/4}} \left(\sin\left(\frac\pi8\right) - i \cos\left(\frac\pi8\right)\right)$

By the residue theorem,

$\displaystyle \int f(z) \, dz = 2\pi i \sum_{\rm poles} \mathrm{Res}\{f(z)\} = \frac{4\pi}{2^{7/4}} \cos\left(\frac\pi8\right)$

We also have

$\displaystyle \cos^2\left(\dfrac\pi8\right) = \dfrac{1 + \cos\left(\frac\pi4\right)}2 = \dfrac{2 + \sqrt2}4 \implies \cos\left(\frac\pi8\right) = \dfrac{\sqrt{2+\sqrt2}}2$

Then the remaining integral is

$\displaystyle \int_0^\infty \frac{t^2+2}{t^4+2t^2+2} \, dt = \frac{4\pi}{2^{7/4}} \cos\left(\frac\pi8\right) = \sqrt{\frac12 + \frac1{\sqrt2}} \, \pi$

It follows that

$\displaystyle \int_0^{\pi/4} \sqrt{\tan(x) - \tan^2(x)} \, dx = \boxed{\left(\sqrt{\frac12 + \frac1{\sqrt2}} - 1\right) \pi}$