∫_0 ^π (1/(a^2 −2a cosx + 1))dx (a<1) is


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Question Number 101585 by Rohit@Thakur last updated on 03/Jul/20

$\int_{0}^{π} \frac{1}{a^{2} - 2 a c o s x + 1} d x (a < 1) i s$
	Answered by mathmax by abdo last updated on 03/Jul/20

	$f (a) = \int_{0}^{π} \frac{dx}{a^{2} - 2 acosx + 1} changement \tan (\frac{x}{2}) = t give$ $f (a) = \int_{0}^{\infty} \frac{2 dt}{(1 + t^{2}) (a^{2} - 2 a \times \frac{1 - t^{2}}{1 + t^{2}} + 1)} = \int_{0}^{\infty} \frac{2 dt}{a^{2} + a^{2} t^{2} - 2 a + {2 at}^{2} + 1 + t^{2}}$ $= \int_{0}^{\infty} \frac{2 dt}{(a^{2} + 2 a + 1) t^{2} + a^{2} - 2 a + 1} = \frac{2}{{(a + 1)}^{2}} \int_{0}^{\infty} \frac{dt}{t^{2} + \frac{{(a - 1)}^{2}}{{(a + 1)}^{2}}}$ $= \frac{2}{{(a + 1)}^{2}} \int_{0}^{\infty} \frac{dt}{t^{2} + {(\frac{a - 1}{a + 1})}^{2}} =_{t = \frac{1 - a}{1 + a} u} \frac{2}{{(a + 1)}^{2}} \int_{0}^{\infty} \frac{1}{{(\frac{1 - a}{1 + a})}^{2} (1 + u^{2})} \times \frac{1 - a}{1 + a} du$ $= \frac{2}{{(1 + a)}^{2}} \times \frac{{(1 + a)}^{2}}{{(1 - a)}^{2}} \times \frac{1 - a}{1 + a} \int_{0}^{\infty} \frac{du}{1 + u^{2}}$ $= \frac{2}{1 - a^{2}} \times \frac{π}{2} = \frac{π}{1 - a^{2}} \Rightarrow f (a) = \frac{π}{1 - a^{2}}$
	Commented byRohit@Thakur last updated on 04/Jul/20

	$T h a n k y o u s i r$
	Commented bymathmax by abdo last updated on 04/Jul/20

	$you are welcome sir .$
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