2022-01-05

Evaluate

Louis Page

It is not difficult to check that for any $b>1$ we have:
$J\left(b\right)={\int }_{0}^{2\pi }\frac{d\theta }{b+\mathrm{cos}\theta }=4{\int }_{0}^{\frac{\pi }{2}}\frac{d\varphi }{b+\mathrm{cos}\left(2\varphi \right)}$
$=4{\int }_{0}^{+\mathrm{\infty }}\frac{dt}{\left(1+{t}^{2}\right)\left(b-1+2{\mathrm{cos}}^{2}\left(\mathrm{arctan}t\right)\right)}=\frac{2\pi }{\sqrt{{b}^{2}-1}}$
hence it follows that:
$-{J}^{\prime }\left(b\right)={\int }_{0}^{2\pi }\frac{d\theta }{{\left(b+\mathrm{cos}\theta \right)}^{2}}=\frac{2b\pi }{{\left({b}^{2}-1\right)}^{\frac{3}{2}}}$
and by taking $a=\frac{1}{b}$ we get:

Cheryl King

We can actually evaluate this integral indefinitely.
This is the most basic thing I could come up with,
Consider
$f\left(x\right)=\frac{\mathrm{sin}x}{1+a\mathrm{cos}x}$
$⇒{f}^{\prime }\left(x\right)=\frac{1+\mathrm{cos}x}{{\left(1+a\mathrm{cos}x\right)}^{2}}$
Now integrate both sides,
$⇒f\left(x\right)=\int \frac{\left(\mathrm{cos}x+a\right)dx}{{\left(1+a\mathrm{cos}x\right)}^{2}}$
Now integrate both sides,
$⇒f\left(x\right)=\int \frac{\left(\mathrm{cos}x+a\right)dx}{{\left(1+a\mathrm{cos}x\right)}^{2}}$
$⇒f\left(x\right)=\frac{1}{a}\int \frac{\left(a\mathrm{cos}x+1\right)dx}{{\left(1+a\mathrm{cos}x\right)}^{2}}+\frac{{a}^{2}-1}{a}\int \frac{dx}{{\left(1+a\mathrm{cos}x\right)}^{2}}$
$⇒\frac{\mathrm{sin}x}{1+a\mathrm{cos}x}=\frac{1}{a}\int \frac{dx}{1+a\mathrm{cos}x}+\frac{{a}^{2}-1}{a}\cdot I$
Now, 'I' is the integral we wanted to evaluate and the only problem left is
$\int \frac{dx}{1+a\mathrm{cos}x}$
which is easily calculated by putting $\mathrm{cos}x=\frac{1-\frac{{\mathrm{tan}}^{2}x}{2}}{1+\frac{{\mathrm{tan}}^{2}x}{2}}$
$\int \frac{dx}{1+a\mathrm{cos}x}=\frac{2}{1-a}\sqrt{\frac{1-a}{1+a}}\mathrm{arctan}\left(t\sqrt{\frac{1-a}{1+a}}\right)+C$
Where $t=\frac{\mathrm{tan}x}{2}$

star233

${\int }_{0}^{2\pi }\frac{d\theta }{\left[1+a\mathrm{cos}\left(\theta \right){\right]}^{2}}={\int }_{-\pi }^{\pi }\frac{d\theta }{\left[1-a\mathrm{cos}\left(\theta \right){\right]}^{2}}=-2\frac{d}{db}{\int }_{0}^{\pi }\frac{d\theta }{b-a\mathrm{cos}\left(\theta \right)}{|}_{b=1}\phantom{\rule{0ex}{0ex}}=-2\frac{d}{db}\left[{\int }_{0}^{\pi /2}\frac{d\theta }{b-a\mathrm{cos}\left(\theta \right)}+{\int }_{0}^{\pi /2}\frac{d\theta }{b+a\mathrm{cos}\left(\theta \right)}{\right]}_{b=1}\phantom{\rule{0ex}{0ex}}=-4\frac{d}{db}\left[b{\int }_{0}^{\pi /2}\frac{d\theta }{{b}^{2}-{a}^{2}{\mathrm{cos}}^{2}\left(\theta \right)}{\right]}_{b=1}\phantom{\rule{0ex}{0ex}}=-4\frac{d}{db}\left[b{\int }_{0}^{\pi /2}\frac{{\mathrm{sec}}^{2}\left(\theta \right)d\theta }{{b}^{2}{\mathrm{tan}}^{2}\left(\theta \right)+{b}^{2}-{a}^{2}}{\right]}_{b=1}\phantom{\rule{0ex}{0ex}}=-4\frac{d}{db}\left(b{\int }_{0}^{\mathrm{\infty }}\frac{dt}{{b}^{2}{t}^{2}+{b}^{2}-{a}^{2}}{\right)}_{b=1}\phantom{\rule{0ex}{0ex}}=-4\frac{d}{db}\left(\frac{1}{\sqrt{{b}^{2}-{a}^{2}}}{\int }_{0}^{\mathrm{\infty }}\frac{dt}{{t}^{2}+1}{\right)}_{b=1}\phantom{\rule{0ex}{0ex}}=-2\pi \frac{d}{db}\left(\frac{1}{\sqrt{{b}^{2}-{a}^{2}}}\right){|}_{b=1}=\frac{2\pi b}{\left({b}^{2}-{a}^{2}{\right)}^{3/2}}{|}_{b=1}\phantom{\rule{0ex}{0ex}}=\frac{2\pi }{\left(1-{a}^{2}{\right)}^{3/2}}$