Evaluate \int_0^{2\pi}\frac{1}{(1+a\cos\theta)^2}d\theta,\ 0\leq a<1

Adela Brown 2022-01-05 Answered
Evaluate
\(\displaystyle{\int_{{0}}^{{{2}\pi}}}{\frac{{{1}}}{{{\left({1}+{a}{\cos{\theta}}\right)}^{{2}}}}}{d}\theta,\ {0}\leq{a}{ < }{1}\)

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Expert Answer

Louis Page
Answered 2022-01-06 Author has 3716 answers
It is not difficult to check that for any \(\displaystyle{b}{>}{1}\) we have:
\(\displaystyle{J}{\left({b}\right)}={\int_{{0}}^{{{2}\pi}}}{\frac{{{d}\theta}}{{{b}+{\cos{\theta}}}}}={4}{\int_{{0}}^{{\frac{\pi}{{2}}}}}{\frac{{{d}\phi}}{{{b}+{\cos{{\left({2}\phi\right)}}}}}}\)
\(\displaystyle={4}{\int_{{0}}^{{+\infty}}}{\frac{{{\left.{d}{t}\right.}}}{{{\left({1}+{t}^{{2}}\right)}{\left({b}-{1}+{2}{{\cos}^{{2}}{\left({\arctan{{t}}}\right)}}\right)}}}}={\frac{{{2}\pi}}{{\sqrt{{{b}^{{2}}-{1}}}}}}\)
hence it follows that:
\(\displaystyle-{J}'{\left({b}\right)}={\int_{{0}}^{{{2}\pi}}}{\frac{{{d}\theta}}{{{\left({b}+{\cos{\theta}}\right)}^{{2}}}}}={\frac{{{2}{b}\pi}}{{{\left({b}^{{2}}-{1}\right)}^{{\frac{{3}}{{2}}}}}}}\)
and by taking \(\displaystyle{a}={\frac{{{1}}}{{{b}}}}\) we get:
\(\displaystyle\forall{a}\in{\left({0},{1}\right)},\ {\int_{{0}}^{{{2}\pi}}}{\frac{{{d}\theta}}{{{\left({1}+{a}{\cos{\theta}}\right)}^{{2}}}}}={\frac{{{2}}}{{{\left({1}-{a}^{{2}}\right)}^{{\frac{{3}}{{2}}}}}}}\)
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Cheryl King
Answered 2022-01-07 Author has 4943 answers
We can actually evaluate this integral indefinitely.
This is the most basic thing I could come up with,
Consider
\(\displaystyle{f{{\left({x}\right)}}}={\frac{{{\sin{{x}}}}}{{{1}+{a}{\cos{{x}}}}}}\)
\(\displaystyle\Rightarrow{f}'{\left({x}\right)}={\frac{{{1}+{\cos{{x}}}}}{{{\left({1}+{a}{\cos{{x}}}\right)}^{{2}}}}}\)
Now integrate both sides,
\(\displaystyle\Rightarrow{f{{\left({x}\right)}}}=\int{\frac{{{\left({\cos{{x}}}+{a}\right)}{\left.{d}{x}\right.}}}{{{\left({1}+{a}{\cos{{x}}}\right)}^{{2}}}}}\)
Now integrate both sides,
\(\displaystyle\Rightarrow{f{{\left({x}\right)}}}=\int{\frac{{{\left({\cos{{x}}}+{a}\right)}{\left.{d}{x}\right.}}}{{{\left({1}+{a}{\cos{{x}}}\right)}^{{2}}}}}\)
\(\displaystyle\Rightarrow{f{{\left({x}\right)}}}={\frac{{{1}}}{{{a}}}}\int{\frac{{{\left({a}{\cos{{x}}}+{1}\right)}{\left.{d}{x}\right.}}}{{{\left({1}+{a}{\cos{{x}}}\right)}^{{2}}}}}+{\frac{{{a}^{{2}}-{1}}}{{{a}}}}\int{\frac{{{\left.{d}{x}\right.}}}{{{\left({1}+{a}{\cos{{x}}}\right)}^{{2}}}}}\)
\(\displaystyle\Rightarrow{\frac{{{\sin{{x}}}}}{{{1}+{a}{\cos{{x}}}}}}={\frac{{{1}}}{{{a}}}}\int{\frac{{{\left.{d}{x}\right.}}}{{{1}+{a}{\cos{{x}}}}}}+{\frac{{{a}^{{2}}-{1}}}{{{a}}}}\cdot{I}\)
Now, 'I' is the integral we wanted to evaluate and the only problem left is
\(\displaystyle\int{\frac{{{\left.{d}{x}\right.}}}{{{1}+{a}{\cos{{x}}}}}}\)
which is easily calculated by putting \(\displaystyle{\cos{{x}}}={\frac{{{1}-\frac{{{\tan}^{{2}}{x}}}{{2}}}}{{{1}+\frac{{{\tan}^{{2}}{x}}}{{2}}}}}\)
\(\displaystyle\int{\frac{{{\left.{d}{x}\right.}}}{{{1}+{a}{\cos{{x}}}}}}={\frac{{{2}}}{{{1}-{a}}}}\sqrt{{{\frac{{{1}-{a}}}{{{1}+{a}}}}}}{\arctan{{\left({t}\sqrt{{{\frac{{{1}-{a}}}{{{1}+{a}}}}}}\right)}}}+{C}\)
Where \(\displaystyle{t}=\frac{{\tan{{x}}}}{{2}}\)
0
star233
Answered 2022-01-11 Author has 0 answers

\(\int_0^{2\pi}\frac{d\theta}{[1+a\cos(\theta)]^2}=\int_{-\pi}^\pi\frac{d\theta}{[1-a\cos(\theta)]^2}=-2\frac{d}{db}\int_0^\pi\frac{d\theta}{b-a\cos(\theta)}|_{b=1} \\=-2\frac{d}{db}[\int_0^{\pi/2}\frac{d\theta}{b-a\cos(\theta)}+\int_0^{\pi/2}\frac{d\theta}{b+a\cos(\theta)}]_{b=1} \\=-4\frac{d}{db}[b\int_0^{\pi/2}\frac{d\theta}{b^2-a^2\cos^2(\theta)}]_{b=1} \\=-4\frac{d}{db}[b\int_0^{\pi/2}\frac{\sec^2(\theta)d\theta}{b^2\tan^2(\theta)+b^2-a^2}]_{b=1} \\=-4\frac{d}{db}(b\int_0^\infty\frac{dt}{b^2t^2+b^2-a^2})_{b=1} \\=-4\frac{d}{db}(\frac{1}{\sqrt{b^2-a^2}}\int_0^\infty\frac{dt}{t^2+1})_{b=1} \\=-2\pi\frac{d}{db}(\frac{1}{\sqrt{b^2-a^2}})|_{b=1}=\frac{2\pi b}{(b^2-a^2)^{3/2}}|_{b=1} \\=\frac{2\pi}{(1-a^2)^{3/2}}\)

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