osnomu3

2022-01-04

Which is the easiest way to evaluate ${\int }_{0}^{\frac{\pi }{2}}\left(\sqrt{\mathrm{tan}x}+\sqrt{\mathrm{cot}x}\right)$ ?

### Answer & Explanation

Durst37

${\int }_{0}^{\frac{\pi }{2}}\sqrt{\mathrm{tan}x}dx+\sqrt{\mathrm{cot}x}dx$
$={\int }_{0}^{\frac{\pi }{2}}\frac{\mathrm{sin}x+\mathrm{cos}x}{\sqrt{\mathrm{sin}x\mathrm{cos}x}}dx={\int }_{0}^{\frac{\pi }{2}}\frac{\mathrm{sin}x+\mathrm{cos}x}{\frac{\sqrt{2\mathrm{sin}x\mathrm{cos}x}}{\sqrt{2}}}dx=\sqrt{2}$
${\int }_{0}^{\frac{\pi }{2}}\frac{\mathrm{sin}x+\mathrm{cos}x}{\sqrt{1-{\left(\mathrm{sin}x-\mathrm{cos}x\right)}^{2}}}dx$
$=\sqrt{2}{\int }_{0}^{\frac{\pi }{2}}\frac{\mathrm{sin}x+\mathrm{cos}x}{\sqrt{1-{\left(\mathrm{sin}x-\mathrm{cos}x\right)}^{2}}}dx$
Let
$x\to \frac{\pi }{2}⇒t=\left(\mathrm{sin}x-\mathrm{cos}x\right)\to 1$
$x\to 0⇒t=\left(\mathrm{sin}x-\mathrm{cos}x\right)\to -1$
$\sqrt{2}{\int }_{-1}^{1}\frac{1}{\sqrt{1-{t}^{2}}}dt=\sqrt{2}{\left[{\mathrm{sin}}^{-1}t\right]}_{-1}^{1}=\sqrt{2}\left[\frac{\pi }{2}-\left(-\frac{\pi }{2}\right)\right]$
$=\sqrt{2}\pi$

Mollie Nash

Let $u=\sqrt{\mathrm{tan}\left(x\right)}$. Then ${u}^{2}=\mathrm{tan}\left(x\right)$ and $2udu=\left(1+{\mathrm{tan}}^{2}\left(x\right)\right)dx$. Thus
${\int }_{0}^{\frac{\pi }{2}}\sqrt{\mathrm{tan}\left(x\right)}dx={\int }_{0}^{\mathrm{\infty }}\frac{2{u}^{2}}{1+{u}^{4}}du$
Since $1+{u}^{4}=\left(1+\sqrt{2}u+{u}^{2}\right)\left(1-\sqrt{2}u+{u}^{2}\right)$, partial fraction decomposition applies:
$\frac{2{u}^{2}}{1+{u}^{4}}=\frac{1}{\sqrt{2}}\left(\frac{u}{{u}^{2}-\sqrt{2}u+1}-\frac{u}{{u}^{2}+\sqrt{2}u+1}\right)$
Hence
$\int \frac{2{u}^{2}}{1+{u}^{4}}du=\frac{1}{2\sqrt{2}}\mathrm{log}\left(\frac{{u}^{2}-\sqrt{2}u+1}{{u}^{2}+\sqrt{2}u+1}+\frac{{\mathrm{tan}}^{-1}\left(\sqrt{2}+1\right)-{\mathrm{tan}}^{-1}\left(1-\sqrt{2}u\right)}{\sqrt{2}}$
Applying the fundamental theorem of calculus:
${\int }_{0}^{\frac{\pi }{2}}\sqrt{\mathrm{tan}\left(x\right)}dx=\frac{\pi }{2}$

star233

We will employing the substitution $u=\sqrt{\mathrm{tan}x}:$
${u}^{\prime }=\frac{1+{\mathrm{tan}}^{2}x}{2\sqrt{\mathrm{tan}x}}$
and
$2{\int }_{0}^{\pi /2}\sqrt{\mathrm{tan}x}dx=4{\int }_{0}^{\mathrm{\infty }}\frac{{u}^{2}}{1+{u}^{4}}du=2{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{{u}^{2}}{1+{u}^{4}}du$
The last integral has two poles () in the upper complex half-plane. The corresponding residue are

Thus the value of the integral is
$2{\int }_{0}^{\pi /2}\sqrt{\mathrm{tan}x}dx=-\pi i\left({u}_{1}+{u}_{2}\right)=\sqrt{2}\pi$

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