For 0 < a < 1, the value of the integral

Engineering

Mathematics

Introduction to Definite Integral

Integration by Substitution

Question

For 0 < a < 1, the value of the integral $\int_{0}^{π} \frac{dx}{1 - 2 acos x + a^{2}} is:$

$\frac{π^{2}}{π + a^{2}}$

$\frac{π^{2}}{π - a^{2}}$

$\frac{π}{1 - a^{2}}$

$\frac{π}{1 + a^{2}}$

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Let $I = \int_{0}^{π} \frac{dx}{1 + a^{2} - 2 acos x}$

$I = \frac{1}{2 a} \int_{0}^{π} \frac{dx}{\frac{1 + a^{2}}{2 a} - \cos x}$

Let $b = \frac{1 + a^{2}}{2 a} then b > 1$

$I = \frac{1}{2 a} \int_{0}^{π} \frac{dx}{b - \cos x}$ where

$= \frac{1}{2 a} \int_{0}^{π} \frac{dx}{b - \frac{1 - \tan^{2} x / 2}{1 + \tan^{2} x / 2}}$

$= \frac{1}{2 a} \int_{0}^{π} \frac{\sec^{2} x / 2}{(b - 1) + (b + 1) \tan^{2} \frac{x}{2}} dx$

Let $\tan \frac{x}{2} = t \Rightarrow \frac{1}{2} \sec^{2} \frac{x}{2} dx = dt$

$= \frac{1}{a} \int_{0}^{\infty} \frac{dt}{(b - 1) + (b + 1) t^{2}} = \frac{1}{a (b + 1)} \int_{0}^{\infty} \frac{dt}{(\frac{b - 1}{b + 1}) + t^{2}} = \frac{1}{a (b + 1)} \int_{0}^{\infty} \frac{dt}{(\frac{b - 1}{b + 1}) + t^{2}}$

$= {\frac{1}{a (b + 1)} \frac{1}{\sqrt{\frac{b - 1}{b + 1}}} \tan^{- 1} (\frac{t}{\sqrt{\frac{b - 1}{b + 1}}})|}_{0}^{\infty} = \frac{1}{a \sqrt{b^{2} - 1}} [\tan^{- 1} \infty - \tan^{- 1} 0]$

$= \frac{π}{2 a \sqrt{b^{2} - 1}} = \frac{π}{2 a \sqrt{{(\frac{1 + a^{2}}{2 a})}^{2} - 1}} = \frac{π}{\sqrt{{(1 - a^{2})}^{2}}} = \frac{π}{|1 - a^{2}|} = \frac{π}{1 - a^{2}} (0 < a < 1)$