find-0-pi-2-x-sinx-dx-

Question Number 59279 by Mr X pcx last updated on 07/May/19

f i n d \int_{0}^{\frac{π}{2}} \frac{x}{s i n x} d x

Commented by tanmay last updated on 07/May/19

f (x) = \frac{x}{s i n x}

f (0) u n d e f i n e d

b u t w h e n \lim_{x \to 0} \frac{x}{s i n x} = 1

f (\frac{π}{2}) = \frac{π}{2}

\int_{0}^{\frac{π}{2}} \frac{x}{s i n x} d x \approx \frac{1}{2} [\lim_{x \to 0} \frac{x}{s i n x} + f (\frac{π}{2})] \times \frac{π}{2}

\approx \frac{1}{2} [1 + \frac{π}{2}] \times \frac{π}{2}

\approx \frac{1}{2} [\frac{π}{2} + \frac{π^{2}}{4}]

p l s c h e c k

Commented by maxmathsup by imad last updated on 10/May/19

w e h a v e p r o v e d t h a t x - \frac{x^{3}}{6} ⩽ s i n x ⩽ x f o r x \in] 0, \frac{π}{2}] \Rightarrow

\frac{1}{x} ⩽ \frac{1}{s i n x} ⩽ \frac{1}{x - \frac{x^{3}}{6}} \Rightarrow 1 ⩽ \frac{x}{s i n x} ⩽ \frac{1}{1 - \frac{x^{2}}{6}} \Rightarrow \frac{π}{2} ⩽ \int_{0}^{\frac{π}{2}} \frac{x}{s i n x} d x ⩽ \int_{0}^{\frac{π}{2}} \frac{d x}{1 - \frac{x^{2}}{6}}

\int_{0}^{\frac{π}{2}} \frac{d x}{1 - \frac{x^{2}}{6}} =_{x = \sqrt{6} t} \int_{0}^{\frac{π}{2 \sqrt{6}}} \frac{\sqrt{6} d t}{1 - t^{2}} = \frac{\sqrt{6}}{2} \int_{0}^{\frac{π}{2 \sqrt{6}}} {\frac{1}{1 - t} + \frac{1}{1 + t}} d t

= \frac{\sqrt{6}}{2} {[l n ∣ \frac{1 + t}{1 - t} ∣]}_{0}^{\frac{π}{2 \sqrt{6}}} = \frac{\sqrt{6}}{2} {l n (\frac{1 + \frac{π}{2 \sqrt{6}}}{1 - \frac{π}{2 \sqrt{6}}})} = \frac{\sqrt{6}}{2} l n (\frac{2 \sqrt{6} + π}{2 \sqrt{6} - π}) \Rightarrow

\frac{π}{2} ⩽ \int_{0}^{\frac{π}{2}} \frac{x}{s i n x} d x ⩽ \frac{\sqrt{6}}{2} l n (\frac{2 \sqrt{6} + π}{2 \sqrt{6} - π}) . i f w e p u t α_{0} = \frac{\sqrt{6}}{2} l n (\frac{2 \sqrt{6} + π}{2 \sqrt{6} - π}) w e c a n

t a k e \frac{\frac{π}{2} + α_{0}}{2} = \frac{π}{4} + \frac{α_{0}}{2} a s a p p r o x i m a t e v a l u e f o r t h i s i n t e g r a l .

Commented by maxmathsup by imad last updated on 10/May/19

s i r T a n m a y w h a t s f o r m u l a e y o u h a v e u s e d i n y o u r a n s w e r .

Commented by tanmay last updated on 10/May/19

i h a v e c a l c u l a t e d t h e a r e a \dots

\int_{a}^{b} f (x) d x \approx \frac{1}{2} [f (a) + f (b)] \times (b - a)

Commented by maxmathsup by imad last updated on 10/May/19

t h a n k s s i r .