How to Verify the Complex Integral Equals π/(1+n)?

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The integral $$\int_{0}^{2\pi}\mathrm dt{\sin t\over \sin t+ i\sqrt{n+\cos^2 t}}$$ equals $$\frac{\pi}{1+n}$$ for all values of n except -1. This conclusion is derived using complex analysis techniques, specifically residue theory and contour integration. The discussion emphasizes the importance of recognizing the singularities in the integrand and applying the residue theorem to evaluate the integral effectively.

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How to prove this integral,

$$\int_{0}^{2\pi}\mathrm dt{\sin t\over \sin t+ i\sqrt{n+\cos^2 t}}={\pi\over 1+n}$$

$n \ne -1$

$i=\sqrt{-1}$
 
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Tony said:
How to prove this integral,

$$\int_{0}^{2\pi}\mathrm dt{\sin t\over \sin t+ i\sqrt{n+\cos^2 t}}={\pi\over 1+n}$$

$n \ne -1$

$i=\sqrt{-1}$
Rationalise the fraction: $$\frac{\sin t}{ \sin t+ i\sqrt{n+\cos^2 t}} = \frac{\sin t\bigl( \sin t - i\sqrt{n+\cos^2 t}\bigr)}{1+n}.$$
 

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