find-1-dx-x-2-2xcos-1-with-0-lt-lt-pi-

Question Number 31056 by abdo imad last updated on 02/Mar/18

f i n d \int_{1}^{+ \infty} \frac{d x}{x^{2} - 2 x c o s α + 1} w i t h 0 < α < π .

Commented by abdo imad last updated on 03/Mar/18

I = \int_{1}^{+ \infty} \frac{d x}{x^{2} - 2 x c o s α + {c o s}^{2} α + {s i n}^{2} α}

= \int_{1}^{+ \infty} \frac{d x}{{(x - c o s α)}^{2} + {s i n}^{2} α} t h e c h . x - c o s α = s i n α t g i v e

I = \int_{\frac{1 - c o s α}{s i n α}}^{+ \infty} \frac{s i n α d t}{{s i n}^{2} α (1 + t^{2})} = \frac{1}{s i n α} \int_{t a n (\frac{α}{2})}^{+ \infty} \frac{d t}{1 + t^{2}}

I = \frac{1}{s i n α} {[a r c t a n t]}_{t a n (\frac{α}{2})}^{+ \infty} = \frac{1}{s i n α} (\frac{π}{2} - \frac{α}{2}) \Rightarrow

I = \frac{π - α}{2 s i n α} .