find-approcimstive-value-of-pi-3-pi-2-x-sinx-dx-

Question Number 88414 by abdomathmax last updated on 10/Apr/20

f i n d a p p r o c i m s t i v e v a l u e o f \int_{\frac{π}{3}}^{\frac{π}{2}} \frac{x}{s i n x} d x

Commented by mathmax by abdo last updated on 12/Apr/20

w e h a v e s i n x = x - \frac{x^{3}}{6} + \dots . \Rightarrow x - \frac{x^{3}}{6} ⩽ s i n x ⩽ x \forall x \in [\frac{π}{3}, \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} - \frac{π}{3} ⩽ \int_{\frac{π}{3}}^{\frac{π}{2}} \frac{x}{s i n x} d x ⩽ 6 \int_{\frac{π}{3}}^{\frac{π}{2}} \frac{d x}{6 - x^{2}}

w e h a v e \int_{\frac{π}{3}}^{\frac{π}{2}} \frac{d x}{6 - x^{2}} = \frac{1}{2 \sqrt{6}} \int_{\frac{π}{3}}^{\frac{π}{2}} (\frac{1}{\sqrt{6} - x} + \frac{1}{\sqrt{6} + x}) d x

= \frac{1}{2 \sqrt{6}} {[l n ∣ \frac{\sqrt{6} + x}{\sqrt{6} - x} ∣]}_{\frac{π}{3}}^{\frac{π}{2}} = \frac{1}{2 \sqrt{6}} {l n (\frac{\sqrt{6} + \frac{π}{2}}{\sqrt{6} - \frac{π}{2}}) - l n (\frac{\sqrt{6} + \frac{π}{3}}{\sqrt{6} - \frac{π}{3}})}

= \frac{1}{2 \sqrt{6}} {l n (\frac{2 \sqrt{6} + π}{2 \sqrt{6} - π}) - l n (\frac{3 \sqrt{6} + π}{3 \sqrt{6} - π})} \Rightarrow

\frac{π}{6} ⩽ \int_{\frac{π}{3}}^{\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}) - l n (\frac{3 \sqrt{6} + π}{3 \sqrt{6} - π})} w e c a n t a k e

v_{0} = \frac{π}{2} + \frac{\sqrt{6}}{4} {l n (\frac{2 \sqrt{6} + π}{2 \sqrt{6} - π}) - l n (\frac{3 \sqrt{6} + π}{3 \sqrt{6} - π})} a s a a p p r o x i m s t e v a l u e

Commented by mathmax by abdo last updated on 12/Apr/20

v_{0} = \frac{π}{12} + \frac{\sqrt{6}}{4} {\dots . .}