Harley Ayers

2022-03-25

Computing by substituting $\mathrm{cos}\left\{x\right\}=\frac{{e}^{ix}+{e}^{-ix}}{2}$?

### Answer & Explanation

glikozyd3s68

A fast way is to exploit real methods.
${\int }_{0}^{\frac{\pi }{2}}\frac{dx}{5+4\mathrm{cos}x}\stackrel{x=2z}{=}2{\int }_{0}^{\frac{\pi }{4}}\frac{dz}{1+8{\mathrm{cos}}^{2}z}\stackrel{z=\mathrm{arctan}u}{=}2{\int }_{0}^{1}\frac{du}{9+{u}^{2}}=$
$=\frac{2}{3}\mathrm{arctan}\frac{1}{3}$

Alejandra Hanna

This is exactly the Kepler angle substitution:
$\mathrm{sin}\psi =\frac{\sqrt{1-{e}^{2}}\mathrm{sin}x}{1+e\mathrm{cos}x}$
$\mathrm{cos}\psi =\frac{\mathrm{cos}x+e}{1+e\mathrm{cos}x}$
$d\psi =\frac{\sqrt{1-{e}^{2}}dx}{1+e\mathrm{cos}x}$
For 0${\int }_{0}^{\frac{\pi }{2}}\frac{dx}{5+4\mathrm{cos}x}=\frac{1}{5}\frac{1}{\sqrt{1-{e}^{2}}}{\int }_{0}^{{\mathrm{cos}}^{-1}e}d\psi =\frac{{\mathrm{cos}}^{-1}e}{5\sqrt{1-{e}^{2}}}$
$=\frac{{\mathrm{cos}}^{-1}\left(\frac{4}{5}\right)}{3}=\frac{1}{3}{\mathrm{tan}}^{-1}\left(\frac{3}{4}\right)=\frac{2}{3}{\mathrm{tan}}^{-1}\left(\frac{1}{3}\right)$
With $e=\frac{4}{5}$

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