find-0-pi-dx-2-cosx-

Question Number 29439 by prof Abdo imad last updated on 08/Feb/18

f i n d \int_{0}^{π} \frac{d x}{2 + c o s x} .

Answered by sma3l2996 last updated on 09/Feb/18

t = t a n x / 2 \Rightarrow d x = \frac{2 d t}{1 + t^{2}}

\int_{0}^{π} \frac{d x}{2 + c o s x} = \int_{0}^{+ \infty} \frac{1}{2 + \frac{1 - t^{2}}{1 + t^{2}}} \times \frac{2 d t}{1 + t^{2}}

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

\sqrt{3} u = t \Rightarrow d t = \sqrt{3} d u

\int_{0}^{π / 2} \frac{d x}{2 + c o s x} = \frac{2 \sqrt{3}}{3} \int_{0}^{+ \infty} \frac{d u}{1 + u^{2}} = \frac{2 \sqrt{3}}{3} {[a r c t a n u]}_{0}^{+ \infty}

= \frac{2 \sqrt{3}}{3} \times \frac{π}{2} = \frac{\sqrt{3} π}{3}