calculate-0-pi-2-x-2-sin-2-x-dx-

Question Number 41054 by turbo msup by abdo last updated on 01/Aug/18

c a l c u l a t e \int_{0}^{\frac{π}{2}} \frac{x^{2}}{{s i n}^{2} x} d x .

Commented by maxmathsup by imad last updated on 02/Aug/18

w e h a v e {c o s}^{2} x = \frac{1}{1 + {t a n}^{2} x} \Rightarrow {s i n}^{2} x = 1 - \frac{1}{1 + {t a n}^{2} x} = \frac{{t a n}^{2} x}{1 + {t a n}^{2} x} \Rightarrow

I = \int_{0}^{\frac{π}{2}} \frac{x^{2}}{{s i n}^{2} x} d x = \int_{0}^{\frac{π}{2}} \frac{(1 + {t a n}^{2} x) x^{2}}{{t a n}^{2} x} d x c h a n g e m e n t t a n x = t g i v e

I = \int_{0}^{\infty} \frac{(1 + t^{2}) {(a r c t a n t)}^{2}}{t^{2}} \frac{d t}{1 + t^{2}} = \int_{0}^{\infty} \frac{{(a r c t a n t)}^{2}}{t^{2}} d t b y p a r t s u^{^{'}} = \frac{1}{t^{2}}

a n d v = {(a r c t a n t)}^{2} \Rightarrow I = {[- \frac{1}{t} {(a r c t a n t)}^{2}]}_{0}^{+ \infty} - \int_{0}^{\infty} - \frac{1}{t} \frac{2 a r c t a n t}{1 + t^{2}} d t

= 2 \int_{0}^{\infty} \frac{a r c t a n t}{t (1 + t^{2})} d t l e t i n t r o d u c e t h e p a r a m e t r i c f u n c t i o n

f (x) = \int_{0}^{\infty} \frac{a r c t a n (x t)}{t (1 + t^{2})} d t w e h a v e

f^{^{'}} (x) = \int_{0}^{\infty} \frac{t}{(1 + x^{2} t^{2}) t (1 + t^{2})} d t = \int_{0}^{\infty} \frac{d t}{(1 + t^{2}) (1 + x^{2} t^{2})} f o r x^{2} \neq 1 w e h a v e

f^{^{'}} (x) = \int_{0}^{\infty} \frac{1}{x^{2} - 1} (\frac{1}{1 + t^{2}} - \frac{1}{1 + x^{2} t^{2}}) d t = \frac{1}{x^{2} - 1} \int_{0}^{\infty} \frac{d t}{1 + t^{2}} - \frac{1}{x^{2} - 1} \int_{0}^{\infty} \frac{d t}{1 + x^{2} t^{2}} b u t

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

\int_{0}^{\infty} \frac{d t}{1 + x^{2} t^{2}} d t =_{x t = u} \int_{0}^{\infty} \frac{1}{1 + u^{2}} \frac{d u}{x} = \frac{1}{x} {[a r c t a n (x t)]}_{0}^{+ \infty} \Rightarrow

f^{^{'}} (x) = \frac{π}{2 (x^{2} - 1)} - \frac{π}{2 x (x^{2} - 1)} = \frac{π}{2 (x^{2} - 1)} {1 - \frac{1}{x}} = \frac{π (x - 1)}{2 x (x^{2} - 1)}

= \frac{π}{2 x (x + 1)} \Rightarrow f (x) = \frac{π}{2} \int_{.}^{x} \frac{d t}{t (t + 1)} = \frac{π}{2} \int_{.}^{x} (\frac{1}{t} - \frac{1}{t + 1}) d t

= \frac{π}{2} l n ∣ \frac{x}{x + 1} ∣ + l w e s e e t h a t l = {l i m}_{x \to + \infty} f (x) \Rightarrow

f (x) = \frac{π}{2} l n ∣ \frac{x}{x + 1} ∣ + l \Rightarrow I = 2 f (1) = 2 {\frac{π}{2} l n (\frac{1}{2}) + l}

= 2 {- \frac{π}{2} l n (2) + l} \Rightarrow I = 2 l - π l n (2) w i t h l = {l i m}_{x \to + \infty} f (x) .