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Can someone help me to evaluate this integral? $$\int \limits _0 ^\beta \frac{\sin\theta}r\,d\theta$$

I need it in a physics exercise. See the image here Given a circle (as shown in diagram), a point $P$ and angle $\theta$ made with the axis.

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  • $\begingroup$ you have to write r as a function of $\theta$ $\endgroup$ – Math chiller Aug 15 '16 at 17:12
  • $\begingroup$ it's also unclear what the integral is trying to evaluate $\endgroup$ – Math chiller Aug 15 '16 at 17:13
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Supposing the circle to be of radius $R$ and its center to be $(0,0)$ and $P=(-a,0)$, we have the point where the ray intersects the circle at $(R\cos\eta,R\sin\eta)$ and so \begin{align} \sin\theta & = \frac{R\sin\eta}{\sqrt{(a+R\cos\eta)^2 + R\sin^2\eta}} = \frac{R\sin\eta}{\sqrt{R^2+a^2 + 2aR\cos\eta}}, \\[10pt] r & = \sqrt{(a+R\cos\eta)^2 + R\sin^2\eta} = \sqrt{R^2+a^2 + 2aR\cos\eta}, \\[15pt] \text{so } \frac{\sin\theta} r & = \frac{R\sin\eta}{R^2+a^2 + 2aR\cos\eta}, \\[10pt] \text{and } d\theta & = \frac{R^2+aR\cos\eta}{R^2+a^2 + 2aR\cos\eta} \, d\eta. \end{align} It's a rational function of sine and cosine, so if all else fails, the standard tangent half-angle substitution will handle it.

PS: Maybe I should be explicit about the derivation of $d\theta$. We have $$ \tan\theta = \frac{\text{rise}}{\text{run}} = \frac{R\cos\eta}{a+R\sin\eta} $$ so $$ \theta = \arctan \frac{R\cos\eta}{a+R\sin\eta} $$ and then differentiate.

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