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Wolfram Alpha indicates the following solution form:-

$$ \int_0^{2\pi} \frac{\sin x}{1 + A \sin x} dx = (1/A)\left( x - \frac{2 \tan^{-1} \left( \frac{A + \tan{(x/2)}}{\sqrt{(1-A^2)}}\right)}{\sqrt{(1-A^2)}} + constant\right)^{2\pi}_0 $$

My thinking is that substituting for $x$ by $2 \pi$ and by $0$ the corresponding values of $\tan(x/2)$ are the same, namely $\tan(\pi) = \tan(0) = 0$. It follows therefore that the fractional term:- $$ \frac{2 \tan^{-1} \left( \frac{A + \tan{(x/2)}}{\sqrt{(1-A^2)}}\right)}{\sqrt{(1-A^2)}} $$ has the same value ( let us call it $Q$ ) for $x=2\pi$ and $x=0.$ Therefore the definite integral becomes

$$ \int_0^{2\pi} \frac{\sin x}{1 + A \sin x} dx = (1/A)\left( (2\pi - Q + constant) - (0 - Q + constant) \right) = 2\pi/A. $$

For $A = 0.2$ we derive the result $$ \int_0^{2\pi} \frac{\sin x}{1 + 0.2 \sin x} dx = 10 \pi. $$

However Wolfram Alpha gives the answer as $-0.64782$ (approx.) which is reasonable looking at the graph.

So what is wrong with my reasoning? ( I realize that the $\tan$ function is not continuous over the range $0,2\pi$ but I don't know what that implies for the analysis of the definite integral ).

EDIT I have accepted @abel's alternative method for solving the integral.

I have posted a related question here which asks why the approach which I followed is incorrect.

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steveOw
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1 Answers1

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$$\int_0^{2\pi}\frac{\sin x}{1+a\sin x }\, dx = \int_0^{2\pi}\left(\frac 1 a-\frac{1}{a(1+a\sin x)}\right)\, dx=\frac{2\pi}a-\frac1a \int_0^{2\pi}\frac{1}{1+a\sin x }\, dx =\frac{2\pi}a-\frac Ja $$

now consider $$\begin{align}J &= \int_0^{2\pi}\frac{1}{1+a\sin x }\, dx\\ &=\int_0^{\pi}\frac{1}{1+a\sin x }\, dx + \int_\pi^{2\pi}\frac{1}{1+a\sin x }\, dx \\ &=\int_0^{\pi}\frac{1}{1+a\sin x }\, dx + \int_0^\pi\frac{1}{1-a\sin x }\, dx \\ &=2\int_0^{\pi/2}\frac{1}{1+a\sin x }\, dx + 2\int_0^{\pi/2}\frac{1}{1-a\sin x }\, dx \\ &= 2f(a) + 2f(-a)\tag 1\end{align}$$

where $$f(a) = \int_0^{\pi/2}\frac{1}{1+a\sin x }\, dx,\quad u = \tan(x/2), x = 2\tan^{-1}u, dx = \frac{2du}{1+u^2} $$ therefore $$\begin{align}f(a) &=2\int_0^1\frac{du}{1+2au+u^2} \\ &= 2\int_0^1\frac{du}{(u+a)^2 + 1- a^2}\\&=\frac{2}{\sqrt{1-a^2}} \tan^{-1}\left(\frac{u+a}{\sqrt{1-a^2}}\right)\big|_0^1\\ &=\frac{2}{\sqrt{1-a^2}}\left(\tan^{-1}\left(\frac{1+a}{\sqrt{1-a^2}}\right) - \tan^{-1}\left(\frac{a}{\sqrt{1-a^2}}\right)\right)\\ &=\frac{2}{\sqrt{1-a^2}}\left(\tan^{-1}\left(\sqrt{\frac{1+a}{1-a}}\right) - \tan^{-1}\left(\frac{a}{\sqrt{1-a^2}}\right)\right)\\ f(a) + f(-a) &= \frac{2}{\sqrt{1-a^2}}\left(\tan^{-1}\left(\sqrt{\frac{1+a}{1-a}}\right) + \tan^{-1}\left(\sqrt{\frac{1-a}{1+a}}\right)\right)\\ &=\frac{\pi}{\sqrt{1-a^2}}\end{align}$$

finally, $$\int_0^{2\pi}\frac{\sin x}{1+a\sin x }\, dx = \frac{2\pi}a-\frac{2\pi}{a\sqrt{1-a^2}}=\frac{2\pi(\sqrt{1-a^2} - 1)}{a\sqrt{1-a^2}}$$

Mark Viola
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abel
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