find-0-pi-dx-a-bcosx-2-with-a-gt-b-gt-0-then-give-the-value-of-0-pi-dx-2-cosx-2-

Question Number 31096 by abdo imad last updated on 02/Mar/18

f i n d \int_{0}^{π} \frac{d x}{{(a + b c o s x)}^{2}} w i t h a > b > 0 t h e n g i v e t h e

v a l u e o f \int_{0}^{π} \frac{d x}{{(2 + c o s x)}^{2}}

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

l e t i n t r o d u c e t h e f u n c t i o n f (a) = \int_{0}^{π} \frac{d x}{a + b c o s x} w e h a v e

f^{^{'}} (a) = - \int_{0}^{π} \frac{d x}{{(a + b c o s x)}^{2}} \Rightarrow \int_{0}^{π} \frac{d x}{{(a + b c o s x)}^{2}} = - f^{^{'}} (a) l e t

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

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

= \int_{0}^{\infty} \frac{2 d t}{a + b + (a - b) t^{2}} = \int_{0}^{\infty} \frac{2 d t}{(a + b) (1 + \frac{a - b}{a + b} t^{2})} t h e n

w e u s e t h e c h . \sqrt{\frac{a - b}{a + b}} t = u \Rightarrow

f (a) = \frac{1}{a + b} \int_{0}^{\infty} \frac{2}{(1 + u^{2})} \sqrt{\frac{a + b}{a - b}} d u

= \frac{2}{\sqrt{a^{2} - b^{2}}} \int_{0}^{\infty} \frac{d u}{1 + u^{2}} = \frac{π}{2} \frac{2}{\sqrt{a^{2} - b^{2}}} = \frac{π}{\sqrt{a^{2} - b^{2}}}

f (a) = π {(a^{2} - b^{2})}^{- \frac{1}{2}} \Rightarrow f^{^{'}} (a) = \frac{- π}{2} (2 a) {(a^{2} - b^{2})}^{- \frac{3}{2}}

= \frac{- π a}{(a^{2} - b^{2}) \sqrt{a^{2} - b^{2}}} \Rightarrow \int_{0}^{π} \frac{d x}{{(a + b c o s x)}^{2}} = \frac{π a}{(a^{2} - b^{2}) \sqrt{a^{2} - b^{2}}}

2) l e t t a k e a = 2 a n d b = 1 \Rightarrow

\int_{0}^{π} \frac{d x}{{(2 + c o s x)}^{2}} = \frac{2 π}{(2^{2} - 1^{2}) \sqrt{2^{2} - 1^{2}}} = \frac{2 π}{3 \sqrt{3}} .