gaiaecologicaq2

2022-07-01

How do you calculate
${\int }_{0}^{5\pi /2}\frac{dx}{2+\mathrm{cos}x}$
I tried all available substitutions including tangent half angle, but all these substitutions do not distinguish between $\pi /2$ and $5\pi /2$. I tried splitting up the integral into two parts, with one part from 0 to $2\pi$, but again this had problems as 0 and $2\pi$ were basically "the same"

1s1zubv

Expert

Substitute (Weierstrass):
$t=\mathrm{tan}\frac{x}{2}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\mathrm{cos}x=\frac{1-{t}^{2}}{1+{t}^{2}}\phantom{\rule{thickmathspace}{0ex}},\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}dx=\frac{2}{1+{t}^{2}}dt\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}$
by the period of cosx (and also of tanx , in fact), we get
${\int }_{0}^{5\pi /2}={\int }_{0}^{\pi }\frac{dx}{2+\mathrm{cos}x}+{\int }_{\pi }^{2\pi }\frac{dx}{2+\mathrm{cos}x}+{\int }_{2\pi }^{5\pi /2}\frac{dx}{2+\mathrm{cos}x}=$
$={\int }_{0}^{\mathrm{\infty }}\frac{2dt}{\left(1+{t}^{2}\right)\left(2+\frac{1-{t}^{2}}{1+{t}^{2}}\right)}+{\int }_{-\mathrm{\infty }}^{0}\frac{2dt}{\left(1+{t}^{2}\right)\left(2+\frac{1-{t}^{2}}{1+{t}^{2}}\right)}+{\int }_{0}^{1}\frac{2dt}{\left(1+{t}^{2}\right)\left(2+\frac{1-{t}^{2}}{1+{t}^{2}}\right)}=$
$=2{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{dt}{3+{t}^{2}}+2{\int }_{0}^{1}\frac{dt}{3+{t}^{2}}=\frac{2}{\sqrt{3}}\left({\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{\frac{1}{\sqrt{3}}dt}{1+{\left(\frac{t}{\sqrt{3}}\right)}^{2}}\right)+\frac{2}{\sqrt{3}}\left({\int }_{0}^{1}\frac{\frac{1}{\sqrt{3}}dt}{1+{\left(\frac{t}{\sqrt{3}}\right)}^{2}}\right)=$
$={\frac{2}{\sqrt{3}}\mathrm{arctan}\frac{t}{\sqrt{3}}|}_{-\mathrm{\infty }}^{\mathrm{\infty }}+{\frac{2}{\sqrt{3}}\mathrm{arctan}\frac{t}{\sqrt{3}}|}_{0}^{1}=\frac{2}{\sqrt{3}}\left(\pi +\frac{\pi }{6}\right)=\frac{7\pi }{3\sqrt{3}}$

Do you have a similar question?