calculate-0-pi-2-dx-1-cosx-cos-with-pi-lt-lt-pi-

Question Number 30178 by abdo imad last updated on 17/Feb/18

c a l c u l a t e \int_{0}^{\frac{π}{2}} \frac{d x}{1 + c o s x c o s θ} w i t h - π < θ < π .

Commented by prof Abdo imad last updated on 22/Feb/18

l e t p u t c o s θ = t I = \int_{0}^{\frac{π}{2}} \frac{d x}{1 + t c o s x} t h e c h . t a n (\frac{x}{2}) = u

g i v e I = \int_{0}^{\infty} \frac{1}{1 + t \frac{1 - u^{2}}{1 + u^{2}}} \frac{2 d u}{1 + u^{2}}

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

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

l e t t b e n u s e t h e c h . \sqrt{\frac{1 - t}{1 + t}} u = α \Rightarrow

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

= \frac{2}{\sqrt{1 - t^{2}}} \int_{0}^{\infty} \frac{d α}{1 + α^{2}} = \frac{π}{\sqrt{1 - t^{2}}} = \frac{π}{\sqrt{1 - {c o s}^{2} θ}} \Rightarrow

I = \frac{π}{∣ s i n θ ∣} i f θ \neq 0 a l s o w e m u s t s t u d y t h e c a s e θ = 0

\dots .