find-the-value-of-F-x-0-pi-2-ln-1-x-sin-2-t-sin-2-t-dt-knowing-that-1-lt-x-lt-1-

Question Number 27781 by abdo imad last updated on 14/Jan/18

f i n d t h e v a l u e o f F (x) = \int_{0}^{\frac{π}{2}} \frac{l n (1 + x {s i n}^{2} t)}{{s i n}^{2} t} d t k n o w i n g t h a t

- 1 < x < 1 .

Commented by abdo imad last updated on 18/Jan/18

a f t e r v e r i f y i n g t h a t F i s d e r i v a b l e o n] - 1, 1 [

F^{,^{'}} (x) = \int_{0}^{\frac{π}{2}} \frac{{s i n}^{2} t}{{s i n}^{2} t (1 + x {s i n}^{2} t)} d t

= \int_{0}^{\frac{π}{2}} \frac{d t}{1 + x {s i n}^{2} t} = \int_{0}^{\frac{π}{2}} \frac{d t}{1 + x \frac{1 - c o s (2 t)}{2}}

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

F^{} (x) = \int_{0}^{π} \frac{d θ}{2 + x - x c o s θ} t h e n t h e c h . t a n (\frac{θ}{2}) = t g i v e

F^{} (x) = \int_{0}^{+ \propto} \frac{1}{2 + x - x \frac{1 - t^{2}}{1 + t^{2}}} \frac{2 d t}{1 + t^{2}}

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

= \int_{0}^{+ \propto} \frac{2 d t}{2 + x + (2 + 2 x) t^{2} - x} = \int_{0}^{+ \propto} \frac{d t}{1 + (1 + x) t^{2}} c h . u = \sqrt{1 + x} t

= \int_{0}^{+ \propto} \frac{d u}{\sqrt{1 + x} (1 + u^{2})} = \frac{π}{2 \sqrt{1 + x}}

\Rightarrow F (x) = \frac{π}{2} \int_{0}^{x} \frac{d t}{\sqrt{1 + t}} + λ

F (x) = π \sqrt{1 + x} + λ a n d λ = F (0) - π = - π s o

F (x) = π (\sqrt{1 + x} - 1) .