The-value-of-the-integral-0-pi-1-a-2-2a-cos-x-1-dx-a-lt-1-is-

Question Number 27295 by iy last updated on 04/Jan/18

The value of the integral

\underset{0}{\int^{π}} \frac{1}{a^{2} - 2 a \cos x + 1} d x (a < 1) is

Commented by abdo imad last updated on 04/Jan/18

l e t d o t h e c h a n g e m e n t t a n (\frac{x}{2}) = t

I = \int_{0}^{π} \frac{d x}{a^{2} - 2 a c o s x + 1} = \int_{0}^{\propto} \frac{\frac{d t}{1 + t^{2}}}{a^{2} - 2 a \frac{1 - t^{2}}{1 + t^{2}} + 1}

= \int_{0}^{\propto} \frac{d t}{a^{2} (1 + t^{2}) - 2 a (1 - t^{2}) + 1 + t^{2}}

= \int_{0}^{\propto} \frac{d t}{(a^{2} + 2 a + 1) t^{2} + a^{2} - 2 a + 1}

= \int_{0}^{\propto} \frac{d t}{{(a + 1)}^{2} t^{2} + {(a - 1)}^{2}}

= \int_{0}^{\propto} \frac{d t}{{(a + 1)}^{2} (t^{2} + {(\frac{1 - a}{a + 1})}^{2})} a n d w e d o t h e c h a n g e e n t

t = \frac{1 - a}{a + 1} α \Rightarrow I = \frac{1}{{(a + 1)}^{2}} \int_{0}^{\propto} \frac{\frac{1 - a}{a + 1} d α}{{(\frac{1 - a}{a + 1})}^{2} (1 + α^{2})}

= \frac{1}{{(a + 1)}^{2}} \frac{1 - a}{a + 1} \frac{{(a + 1)}^{2}}{{(1 - a)}^{2}} \frac{π}{2}

= \frac{π}{2 (1 - a^{2})} .