calculate-0-pi-2-x-sinx-dx-

Question Number 40868 by math khazana by abdo last updated on 28/Jul/18

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

Commented by tanmay.chaudhury50@gmail.com last updated on 29/Jul/18

Commented by tanmay.chaudhury50@gmail.com last updated on 29/Jul/18

\frac{Π}{2} = 1.57 s o 0 < \int_{0}^{\frac{Π}{2}} \frac{x}{s i n x} ⩽ 18 p l s c h e c k

Commented by maxmathsup by imad last updated on 29/Jul/18

t h e Q i s c a l c u l a t e \int_{\frac{π}{3}}^{\frac{π}{2}} \frac{x}{s i n x} d x .

Commented by math khazana by abdo last updated on 30/Jul/18

l e t I = \int_{\frac{π}{3}}^{\frac{π}{2}} \frac{x}{s i n x} d x c h a n g e m e n t t a n (\frac{x}{2}) = t g i v e

I = \int_{\frac{1}{\sqrt{3}}}^{1} \frac{2 a r c t a n t}{\frac{2 t}{1 + t^{2}}} \frac{2 d t}{1 + t^{2}} = 2 \int_{\frac{1}{\sqrt{3}}}^{1} \frac{a r c t a n t}{t} 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_{\frac{1}{\sqrt{3}}}^{1} \frac{a r c t a n (x t)}{t} d t w e h a v e

f^{^{'}} (x) = \int_{\frac{1}{\sqrt{3}}}^{1} \frac{t}{t (1 + x^{2} t^{2})} d t = \int_{\frac{1}{\sqrt{3}}}^{1} \frac{d t}{1 + x^{2} t^{2}}

=_{x t = u} \int_{\frac{x}{\sqrt{3}}}^{x} \frac{1}{1 + u^{2}} \frac{d u}{x} = \frac{1}{x} \int_{\frac{x}{\sqrt{3}}}^{x} \frac{d u}{1 + u^{2}}

= \frac{1}{x} {[a r c t a n (u)]}_{\frac{x}{\sqrt{3}}}^{x} = \frac{1}{x} {a r c t a n (x) - a r c t a n (\frac{x}{\sqrt{3}})} \Rightarrow

f (x) = \int_{0}^{x} \frac{a r c t a n (t)}{t} d t - \int_{0}^{x} \frac{a r c t a n (\frac{t}{\sqrt{3}})}{t} d t + λ

λ = f (0) = 0 b u t \int_{0}^{x} \frac{a r c t a n (\frac{t}{\sqrt{3}})}{t} d t

=_{\frac{t}{\sqrt{3}} = u} \int_{0}^{\frac{x}{\sqrt{3}}} \frac{a r c t a n (u)}{u \sqrt{3}} \sqrt{3} d u = \int_{0}^{\frac{x}{\sqrt{3}}} \frac{a r c t a n (t)}{t} d t \Rightarrow

f (x) = \int_{0}^{x} \frac{a r c t a n (t)}{t} d t - \int_{0}^{\frac{x}{\sqrt{3}}} \frac{a r c t a n (t)}{t} d t

I = 2 f (1) = 2 {\int_{0}^{1} \frac{a r c t a n t}{t} d t - \int_{0}^{\frac{1}{\sqrt{3}}} \frac{a r c t a n t}{t} d t}

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

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

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

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

= \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{{(2 n + 1)}^{2}} (s u m k n o w n b y f o u r i e r s e r i e)

\int_{0}^{\frac{1}{\sqrt{3}}} \frac{a r c t a n t}{t} d t = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{(n + 1 +} {[\frac{1}{2 n + 1} t^{2 n + 1}]}_{0}^{\frac{1}{\sqrt{3}}}

= \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{{(2 n + 1)}^{2}} {(\frac{1}{\sqrt{3}})}^{2 n + 1} \dots