find-0-pi-4-x-2-1-cos-x-dx-

Question Number 62001 by maxmathsup by imad last updated on 13/Jun/19

f i n d \int_{0}^{\frac{π}{4}} \frac{x^{2}}{1 - c o s (x)} d x

Commented by maxmathsup by imad last updated on 14/Jun/19

a p p r o x i m a t e v a l u e l e t I = \int_{0}^{\frac{π}{4}} \frac{x^{2}}{1 - c o s x} d x w e h a v e

c o s x = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} x^{2 n}}{(2 n)!} (w i t h r a d i u s R = \infty) \Rightarrow c o s x = 1 - \frac{x^{2}}{2} + \frac{x^{4}}{4!} - \dots \Rightarrow 1 - \frac{x^{2}}{2} ⩽ c o s x ⩽ 1 - \frac{x^{2}}{2} + \frac{x^{4}}{4!} \Rightarrow

- 1 + \frac{x^{2}}{2} - \frac{x^{4}}{4!} ⩽ - c o s x ⩽ - 1 + \frac{x^{2}}{2} \Rightarrow \frac{x^{2}}{2} - \frac{x^{4}}{4!} ⩽ 1 - c o s x ⩽ \frac{x^{2}}{2} \Rightarrow

\frac{2}{x^{2}} ⩽ \frac{1}{1 - c o s x} ⩽ \frac{1}{\frac{x^{2}}{2} - \frac{x^{4}}{4!}} \Rightarrow 2 ⩽ \frac{x^{2}}{1 - c o s x} ⩽ \frac{1}{\frac{1}{2} - \frac{x^{2}}{4!}} \Rightarrow

\int_{0}^{\frac{π}{4}} 2 d x ⩽ \int_{0}^{\frac{π}{4}} \frac{x^{2}}{1 - c o s x} d x ⩽ \int_{0}^{\frac{π}{4}} \frac{d x}{\frac{1}{2} - \frac{x^{2}}{4!}} \Rightarrow \frac{π}{2} ⩽ I ⩽ 2.4! \int_{0}^{\frac{π}{4}} \frac{d x}{4! - 2 x^{2}}

\int_{0}^{\frac{π}{4}} \frac{d x}{4! - 2 x^{2}} = \int_{0}^{\frac{π}{4}} \frac{d x}{24 - 2 x^{2}} = \frac{1}{2} \int_{0}^{\frac{π}{4}} \frac{d x}{12 - x^{2}} = \frac{1}{2} \int_{0}^{\frac{π}{4}} \frac{d x}{(2 \sqrt{3} - x) (2 \sqrt{3} + x)}

= \frac{1}{2} \frac{1}{4 \sqrt{3}} \int_{0}^{\frac{π}{4}} {\frac{1}{2 \sqrt{3} - x} + \frac{1}{2 \sqrt{3} + x}} d x = \frac{1}{8 \sqrt{3}} {[l n ∣ \frac{2 \sqrt{3} + x}{2 \sqrt{3} - x} ∣]}_{0}^{\frac{π}{4}}

= \frac{1}{8 \sqrt{3}} l n ∣ \frac{2 \sqrt{3} + \frac{π}{4}}{2 \sqrt{3} - \frac{π}{4}} ∣ = \frac{1}{8 \sqrt{3}} l n ∣ \frac{8 \sqrt{3} + π}{8 \sqrt{3} - π} ∣ \Rightarrow \frac{π}{2} ⩽ I ⩽ \frac{2.24}{8 \sqrt{3}} l n ∣ \frac{8 \sqrt{3} + π}{8 \sqrt{3} - π} ∣ \Rightarrow

\frac{π}{2} ⩽ I ⩽ \frac{6}{\sqrt{3}} l n ∣ \frac{8 \sqrt{3} + π}{8 \sqrt{3} - π} ∣ l e t v_{0} = \frac{π}{4} + \sqrt{3} l n (\frac{8 \sqrt{3} + π}{8 \sqrt{3} - π})

v_{0} i s a 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 .