approximate-0-pi-2-x-sinx-dx-by-simpsom-method-

Question Number 94285 by mathmax by abdo last updated on 17/May/20

a p p r o x i m a t e \int_{0}^{\frac{π}{2}} \frac{x}{s i n x} d x b y s i m p s o m m e t h o d

Commented by PRITHWISH SEN 2 last updated on 18/May/20

taking 7 ordinates at x = 0, \frac{π}{12}, \frac{π}{6}, \frac{π}{4}, \frac{π}{3}, \frac{5 π}{12}, \frac{π}{2}

here h = \frac{1}{6}

x f (x) = \frac{x}{\sin x}

0 ∽ 1 = y_{0} (considering the limiting value)

\frac{π}{12} 1.011515 = y_{1}

\frac{π}{6} 1.047197 = y_{2}

\frac{π}{4} 1.110720 = y_{3}

\frac{π}{3} 1.209199 = y_{4}

\frac{5 π}{12} 1.355173 = y_{5}

\frac{π}{2} 1.570796 = y_{6}

applying {Simpson}^{'} s one third rule

I_{\frac{1}{3}} = \frac{h}{3} [(y_{0} + y_{6}) + 4 (y_{1} + y_{3} + y_{5}) + 2 (y_{2} + y_{4})]

= \frac{1}{18} {2.570796 + 13.909632 + 4.512792}

= 1.16629 please check .

Commented by mathmax by abdo last updated on 20/May/20

thank you sir .

Commented by PRITHWISH SEN 2 last updated on 20/May/20

you are welcome

Answered by mathmax by abdo last updated on 20/May/20

we have f (x) = \frac{x}{sinx} \Rightarrow \int_{a}^{b} f (x) \sim \frac{b - a}{6} {f (a) + 4 f (\frac{a + b}{2}) + f (b)} \Rightarrow

\Rightarrow \int_{0}^{\frac{π}{2}} \frac{x}{sinx} dx \sim \frac{π}{6} {f (0) + 4 f (\frac{π}{4}) + f (\frac{π}{2})} \Rightarrow

\int_{0}^{\frac{π}{2}} \frac{x}{sinx} dx \sim \frac{π}{6} (1 + 4 \times \frac{π}{4 \times \frac{1}{\sqrt{2}}} + \frac{π}{2}) (1 = \lim_{x \to 0} \frac{x}{sinx})

\int_{0}^{\frac{π}{2}} \frac{x}{sinx} dx \sim \frac{π}{6} (1 + π \sqrt{2} + \frac{π}{2})