calculate-0-pi-dx-1-2cosx-

Question Number 31070 by abdo imad last updated on 02/Mar/18

c a l c u l a t e \int_{0}^{π} \frac{d x}{1 + 2 c o s x} .

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

t h e c h . t a n (\frac{x}{2}) = t g i v e

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

= \int_{0}^{+ \infty} \frac{2 d t}{3 - t^{2}} = \frac{1}{2 \sqrt{3}} \int_{0}^{+ \infty} 2 (\frac{1}{\sqrt{3} - t} + \frac{1}{\sqrt{3} + t}) d t

= \frac{1}{\sqrt{3}} {[l n (\sqrt{3} + t) - l n (\sqrt{3} - t)]}_{0}^{+ \infty} = \frac{1}{\sqrt{3}} {[l n (∣ \frac{\sqrt{3} + t}{\sqrt{3} - t}) ∣]}_{0}^{+ \infty} \Rightarrow

I = 0