find-the-value-of-pi-6-pi-4-x-1-cos-2-x-dxr-

Question Number 41052 by turbo msup by abdo last updated on 01/Aug/18

f i n d t h e v a l u e o f \int_{\frac{π}{6}}^{\frac{π}{4}} \frac{x}{1 + {c o s}^{2} x} d x r

Commented by maxmathsup by imad last updated on 02/Aug/18

l e t A = \int_{\frac{π}{6}}^{\frac{π}{4}} \frac{x}{1 + {c o s}^{2} x} d x w e h a v e A = \int_{\frac{π}{6}}^{\frac{π}{4}} \frac{x}{1 + \frac{1}{1 + {t a n}^{2} x}} d x

= \int_{\frac{π}{6}}^{\frac{π}{4}} \frac{x (1 + {t a n}^{2} x)}{2 + {t a n}^{2} x} d x c h a n g e m e n t t a n x = t g i v e

A = \int_{\frac{1}{\sqrt{3}}}^{1} \frac{a r c t a n t (1 + t^{2})}{2 + t^{2}} \frac{d t}{1 + t^{2}} = \int_{\frac{1}{\sqrt{3}}}^{1} \frac{a r c t a n t}{2 + t^{2}} 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 φ (x) = \int_{\frac{1}{\sqrt{3}}}^{1} \frac{a r c t a n (x t)}{2 + t^{2}} d t w e h a v e

φ^{^{'}} (x) = \int_{\frac{1}{\sqrt{3}}}^{1} \frac{t}{(1 + x^{2} t^{2}) (2 + t^{2})} d t =_{x t = u} \int_{\frac{x}{\sqrt{3}}}^{x} \frac{u}{x (1 + u^{2}) (2 + \frac{u^{2}}{x^{2}})} \frac{d u}{x}

= \int_{\frac{x}{\sqrt{3}}}^{x} \frac{u}{(1 + u^{2}) (2 x^{2} + u^{2})} d u l e t d e c o m p o s e F (u) = \frac{u}{(u^{2} + 1) (u^{2} + 2 x^{2})}

F (u) = \frac{a u + b}{u^{2} + 1} + \frac{c u + d}{u^{2} + 2 x^{2}}

F (- u) = - F (u) \Rightarrow \frac{- a u + b}{u^{2} + 1} + \frac{- c u + d}{u^{2} + 2 x^{2}} = \frac{- a u - b}{u^{2} + 1} + \frac{- c u - d}{u^{2} + 2 x^{2}} \Rightarrow

b = d = 0 \Rightarrow F (u) = \frac{a u}{u^{2} + 1} + \frac{c u}{u^{2} + 2 x^{2}}

{l i m}_{u \to + \infty} u F (u) = 0 = a + c \Rightarrow c = - a \Rightarrow F (u) = \frac{a u}{u^{2} + 1} - \frac{a u}{u^{2} + 2 x^{2}}

F (1) = \frac{1}{2 (1 + 2 x^{2})} = \frac{a}{2} - \frac{a}{1 + 2 x^{2}} \Rightarrow \frac{1}{2} = \frac{1 + 2 x^{2}}{2} a - a \Rightarrow

1 = (1 + 2 x^{2}) a - 2 a = (2 x^{2} - 1) a \Rightarrow a = \frac{1}{2 x^{2} - 1} \Rightarrow

F (u) = \frac{1}{2 x^{2} - 1} {\frac{u}{u^{2} + 1} - \frac{u}{u^{2} + 2 x^{2}}} \Rightarrow φ^{^{'}} (x) = \frac{1}{2 x^{2} - 1} \int_{\frac{x}{\sqrt{3}}}^{x} (\frac{u}{u^{2} + 1} - \frac{u}{u^{2} + 2 x^{2}}) d u

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

= \frac{1}{2 (2 x^{2} - 1)} {l n (\frac{x^{2} + 1}{3 x^{2}}) - l n (\frac{x^{2} + 3}{7 x^{2}})}

= \frac{1}{2 (2 x^{2} - 1)} {l n (x^{2} + 1) - l n (3) - 2 l n ∣ x ∣ - l n (x^{2} + 3) + l n (7) + 2 l n ∣ x ∣}

= \frac{l n (7) - l n (3)}{2 (2 x^{2} - 1)} + \frac{l n (x^{2} + 1)}{2 (2 x^{2} - 1)} - \frac{l n (x^{2} + 3)}{2 (2 x^{2} - 1)} \Rightarrow

φ (x) = \int \frac{l n (7) - l n (3)}{2 (2 x^{2} - 1)} d x + \int \frac{l n (x^{2} + 1)}{2 (2 x^{2} - 1)} d x - \int \frac{l n (x^{2} + 3)}{2 (2 x^{2} - 1)} d x + c

\dots b e c o n t i n u e d \dots .