find-the-value-of-0-pi-2-x-1-cosx-dx-

Question Number 43999 by maxmathsup by imad last updated on 19/Sep/18

f i n d t h e v a l u e o f \int_{0}^{\frac{π}{2}} \frac{x}{\sqrt{1 - c o s x}} d x .

Commented by maxmathsup by imad last updated on 20/Sep/18

l e t I = \int_{0}^{\frac{π}{2}} \frac{x}{\sqrt{1 - c o s x}} d x \Rightarrow I = \int_{0}^{\frac{π}{2}} \frac{x}{\sqrt{2 {s i n}^{2} (\frac{x}{2})}} d x

= \frac{1}{\sqrt{2}} \int_{0}^{\frac{π}{2}} \frac{x}{s i n (\frac{x}{2})} d x =_{\frac{x}{2} = t} \frac{1}{\sqrt{2}} \int_{0}^{\frac{π}{4}} \frac{2 t}{s i n (t)} 2 d t = \frac{4}{\sqrt{2}} \int_{0}^{\frac{π}{4}} \frac{t}{s i n t} d t

c h a n g e m e n t t a n (\frac{t}{2}) = u g i v e

I = \frac{4}{\sqrt{2}} \int_{0}^{\sqrt{2} - 1} \frac{2 a r c t a n (u)}{\frac{2 u}{1 + u^{2}}} \frac{2 d u}{1 + u^{2}} = 4 \sqrt{2} \int_{0}^{\sqrt{2} - 1} \frac{a r c t a n (u)}{u} d u

l e t f (x) = \int_{0}^{\sqrt{2} - 1} \frac{a r c t a n (x u)}{u} d u \Rightarrow I = 4 \sqrt{2} f (1)

w e h a v e f^{^{'}} (x) = \int_{0}^{\sqrt{2} - 1} \frac{d u}{1 + x^{2} u^{2}} =_{x u = α} \int_{0}^{x (\sqrt{2} - 1)} \frac{1}{1 + α^{2}} \frac{d α}{x}

= \frac{1}{x} \int_{0}^{x (\sqrt{2} - 1)} \frac{d α}{1 + α^{2}} = \frac{1}{x} {[a r c t a n (α)]}_{0}^{x (\sqrt{2} - 1)}

= \frac{a r c t a n {x (\sqrt{2} - 1)}}{x} \Rightarrow f (x) = \int_{0}^{x} \frac{a r c t a n {t (\sqrt{2} - 1)}}{t} d t + k

k = f (0) = 0 \Rightarrow f (x) = \int_{0}^{x} \frac{a r c t a n {t (\sqrt{2} - 1)}}{t} d t \Rightarrow

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

{a r c t a n}^{^{'}} (x) = \frac{1}{1 + x^{2}} = \sum_{n = 0}^{\infty} {(- 1)}^{n} x^{2 n} \Rightarrow a r c t a n x = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{2 n + 1} x^{2 n + 1} \Rightarrow

a r c t a n {t (\sqrt{2} - 1)} = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{2 n + 1} t^{2 n + 1} {(\sqrt{2} - 1)}^{2 n + 1} \Rightarrow

f (1) = \sum_{n = 0}^{\infty} \int_{0}^{1} \frac{{(- 1)}^{n} {(\sqrt{2} - 1)}^{2 n + 1}}{2 n + 1} t^{2 n} d t

= \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} {(\sqrt{2} - 1)}^{2 n + 1}}{{(2 n + 1)}^{2}} a n d I = 4 \sqrt{2} f (1) \dots