calculate ∫_0 ^(π/2) (x/(sinx))dx .


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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
$let I = ∫_(π/3) ^(π/2) (x/(sinx))dx changement tan((x/2))=t give I = ∫_(1/(√3)) ^1 ((2arctant)/((2t)/(1+t^2 ))) ((2dt)/(1+t^2 )) =2 ∫_(1/(√3)) ^1 ((arctant)/t) dt let introduce the parametric function f(x) = ∫_(1/(√3)) ^1 ((arctan(xt))/t)dt we have f^′ (x)= ∫_(1/(√3)) ^1 (t/(t(1+x^2 t^2 )))dt =∫_(1/(√3)) ^1 (dt/(1+x^2 t^2 )) =_(xt=u) ∫_(x/(√3)) ^x (1/(1+u^2 )) (du/x) =(1/x) ∫_(x/(√3)) ^x (du/(1+u^2 )) =(1/x)[arctan(u)]_(x/(√3)) ^x =(1/x){arctan(x)−arctan((x/(√3)))}⇒ f(x)= ∫_0 ^x ((arctan(t))/t)dt −∫_0 ^x ((arctan((t/(√3))))/t)dt +λ λ=f(0)=0 but ∫_0 ^x ((arctan((t/(√3))))/t)dt =_((t/(√3))=u) ∫_0 ^(x/(√3)) ((arctan(u))/(u(√3))) (√3)du =∫_0 ^(x/(√3)) ((arctan(t))/t)dt⇒ f(x)= ∫_0 ^x ((arctan(t))/t)dt−∫_0 ^(x/(√3)) ((arctan(t))/t)dt I =2f(1) =2{ ∫_0 ^1 ((arctant)/t)dt −∫_0 ^(1/(√3)) ((arctant)/t)dt} we have arctan^′ (t)=(1/(1+t^2 )) =Σ_(n=0) ^∞ (−1)^n t^(2n) ⇒ arctant = Σ_(n=0) ^∞ (((−1)^n )/(2n+1))t^(2n+1) ⇒ ((arctant)/t) =Σ_(n=0) ^∞ (((−1)^n )/(2n+1)) t^(2n) ⇒ ∫_0 ^1 ((arctant)/t)dt =Σ_(n=0) ^∞ (((−1)^n )/(2n+1))∫_0 ^1 t^(2n) dt =Σ_(n=0) ^∞ (((−1)^n )/((2n+1)^2 )) (sum knownby fourier serie) ∫_0 ^(1/(√3)) ((arctant)/t)dt =Σ_(n=0) ^∞ (((−1)^n )/((n+1+))[(1/(2n+1))t^(2n+1) ]_0 ^(1/(√3)) =Σ_(n=0) ^∞ (((−1)^n )/((2n+1)^2 )) ((1/(√3)))^(2n+1) ...$
$l e t I = \int_{\frac{π}{3}}^{\frac{π}{2}} \frac{x}{s i n x} d x c h a n g e m e n t t a n (\frac{x}{2}) = t g i v e$ $I = \int_{\frac{1}{\sqrt{3}}}^{1} \frac{2 a r c t a n t}{\frac{2 t}{1 + t^{2}}} \frac{2 d t}{1 + t^{2}} = 2 \int_{\frac{1}{\sqrt{3}}}^{1} \frac{a r c t a n t}{t} d t$ $l e t i n t r o d u c e t h e p a r a m e t r i c f u n c t i o n$ $f (x) = \int_{\frac{1}{\sqrt{3}}}^{1} \frac{a r c t a n (x t)}{t} d t w e h a v e$ $f^{^{'}} (x) = \int_{\frac{1}{\sqrt{3}}}^{1} \frac{t}{t (1 + x^{2} t^{2})} d t = \int_{\frac{1}{\sqrt{3}}}^{1} \frac{d t}{1 + x^{2} t^{2}}$ $=_{x t = u} \int_{\frac{x}{\sqrt{3}}}^{x} \frac{1}{1 + u^{2}} \frac{d u}{x} = \frac{1}{x} \int_{\frac{x}{\sqrt{3}}}^{x} \frac{d u}{1 + u^{2}}$ $= \frac{1}{x} {[a r c t a n (u)]}_{\frac{x}{\sqrt{3}}}^{x} = \frac{1}{x} {a r c t a n (x) - a r c t a n (\frac{x}{\sqrt{3}})} \Rightarrow$ $f (x) = \int_{0}^{x} \frac{a r c t a n (t)}{t} d t - \int_{0}^{x} \frac{a r c t a n (\frac{t}{\sqrt{3}})}{t} d t + λ$ $λ = f (0) = 0 b u t \int_{0}^{x} \frac{a r c t a n (\frac{t}{\sqrt{3}})}{t} d t$ $=_{\frac{t}{\sqrt{3}} = u} \int_{0}^{\frac{x}{\sqrt{3}}} \frac{a r c t a n (u)}{u \sqrt{3}} \sqrt{3} d u = \int_{0}^{\frac{x}{\sqrt{3}}} \frac{a r c t a n (t)}{t} d t \Rightarrow$ $f (x) = \int_{0}^{x} \frac{a r c t a n (t)}{t} d t - \int_{0}^{\frac{x}{\sqrt{3}}} \frac{a r c t a n (t)}{t} d t$ $I = 2 f (1) = 2 {\int_{0}^{1} \frac{a r c t a n t}{t} d t - \int_{0}^{\frac{1}{\sqrt{3}}} \frac{a r c t a n t}{t} d t}$ $w e h a v e {a r c t a n}^{^{'}} (t) = \frac{1}{1 + t^{2}} = \sum_{n = 0}^{\infty} {(- 1)}^{n} t^{2 n} \Rightarrow$ $a r c t a n t = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{2 n + 1} t^{2 n + 1} \Rightarrow$ $\frac{a r c t a n t}{t} = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{2 n + 1} t^{2 n} \Rightarrow$ $\int_{0}^{1} \frac{a r c t a n t}{t} d t = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{2 n + 1} \int_{0}^{1} t^{2 n} d t$ $= \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{{(2 n + 1)}^{2}} (s u m k n o w n b y f o u r i e r s e r i e)$ $\int_{0}^{\frac{1}{\sqrt{3}}} \frac{a r c t a n t}{t} d t = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{(n + 1 +} {[\frac{1}{2 n + 1} t^{2 n + 1}]}_{0}^{\frac{1}{\sqrt{3}}}$ $= \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{{(2 n + 1)}^{2}} {(\frac{1}{\sqrt{3}})}^{2 n + 1} . . .$
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