find-0-pi-6-cosx-ln-cosx-dx-

Question Number 50420 by Abdo msup. last updated on 16/Dec/18

f i n d \int_{0}^{\frac{π}{6}} c o s x l n (c o s x) d x

Commented by tanmay.chaudhury50@gmail.com last updated on 17/Dec/18

l e t I = \int_{0}^{\frac{π}{6}} c o s x l n (c o s x) d x

∣ I ∣ i s t h e a r e a u n d e r t h e c u r v e i n t h e g i v e n i n t e r v a l

a r e a i s n e g e t i v e b u t c o n s i d e r t h e v a l u e o n l y \dots

Commented by tanmay.chaudhury50@gmail.com last updated on 17/Dec/18

Commented by tanmay.chaudhury50@gmail.com last updated on 17/Dec/18

∣ I ∣< \frac{1}{2} \times \frac{π}{6} \times 0.1246

r e q u i r e d a r e a i s n e g e t i v e v a l u e b u t c o n s i d e r i n g

t h e m a n e t u d e o n l y \dots

∣ I ∣< 0.0326 (a p p r o x)

Commented by maxmathsup by imad last updated on 24/Dec/18

l e t A = \int_{0}^{\frac{π}{6}} c o s x l n (c o s x) d x b y p a r t s u^{^{'}} = c o s x a n d v = l n (c o s x) \Rightarrow

A = {[s i n x l n (c o s x)]}_{0}^{\frac{π}{6}} - \int_{0}^{\frac{π}{6}} s i n x (- \frac{s i n x}{c o s x}) d x

= \frac{1}{2} l n (\frac{\sqrt{3}}{2}) + \int_{0}^{\frac{π}{6}} \frac{{s i n}^{2} x}{c o s x} d x b u t \int_{0}^{\frac{π}{6}} \frac{{s i n}^{2} x}{c o s x} d x = \int_{0}^{\frac{π}{6}} \frac{1 - {c o s}^{2} x}{c o s x} d x

= \int_{0}^{\frac{π}{6}} \frac{d x}{c o s x} - \int_{0}^{\frac{π}{6}} c o s x d x

\int_{0}^{\frac{π}{6}} c o s x d x = {[s i n x]}_{0}^{\frac{π}{6}} = \frac{1}{2}

\int_{0}^{\frac{π}{6}} \frac{1}{c o s x} d x =_{t a n (\frac{x}{2}) = u} \int_{0}^{t a n (\frac{π}{12})} \frac{1}{\frac{1 - u^{2}}{1 + u^{2}}} \frac{2 d u}{1 + u^{2}} = \int_{0}^{t a n (\frac{π}{12})} \frac{2 d u}{(1 - u) (1 + u)}

= \int_{0}^{t a n (\frac{π}{12})} (\frac{1}{1 - u} + \frac{1}{1 + u}) d u = {[l n ∣ \frac{1 + u}{1 - u} ∣]}_{0}^{t a n (\frac{π}{12})} = l n ∣ \frac{1 + t a n (\frac{π}{12})}{1 - t a n (\frac{π}{12})} ∣

{c o s}^{2} (\frac{π}{12}) = \frac{1 + c o s (\frac{π}{6})}{2} = \frac{1 + \frac{\sqrt{3}}{2}}{2} = \frac{2 + \sqrt{3}}{4} \Rightarrow c o s (\frac{π}{12}) = \frac{\sqrt{2 + \sqrt{3}}}{2}

w e f i n d a l s o s i n (\frac{π}{12}) = \frac{\sqrt{2 - \sqrt{3}}}{2} \Rightarrow t a n (\frac{π}{12}) = \frac{\sqrt{2 - \sqrt{3}}}{\sqrt{2 + \sqrt{3}}} = \frac{2 - \sqrt{3}}{\sqrt{4 - 3}} = 2 - \sqrt{3} \Rightarrow

\int_{0}^{\frac{π}{6}} \frac{d x}{c o s x} = l n ∣ \frac{3 - \sqrt{3}}{- 1 + \sqrt{3}} ∣ = l n ∣ \frac{\sqrt{3} (\sqrt{3} - 1)}{\sqrt{3} - 1} ∣= l n (\sqrt{3}) \Rightarrow

A = \frac{1}{4} l n (\frac{3}{2}) - \frac{1}{2} + \frac{1}{2} l n (3) = \frac{3}{4} l n (3) - \frac{1}{4} l n (2) - \frac{1}{2} .

Answered by Smail last updated on 17/Dec/18

b y p a r t s

u = l n (c o s x) \Rightarrow u^{'} = \frac{- s i n x}{c o s x}

v^{'} = c o s x \Rightarrow v = s i n x

I = \int_{0}^{π / 6} c o s x l n (c o s x) d x

= {[s i n x l n (c o s x)]}_{0}^{π / 6} + \int \frac{{s i n}^{2} x}{c o s x} d x

= \frac{1}{2} l n (\frac{\sqrt{3}}{2}) + \int_{0}^{π / 6} \frac{1 - {c o s}^{2} x}{c o s x} d x

= \frac{1}{2} l n (\frac{\sqrt{3}}{2}) + \int_{0}^{π / 6} \frac{d x}{c o s x} - \int_{0}^{π / 6} c o s x d x

l e t t = t a n (x / 2) \Rightarrow d x = \frac{2 d t}{1 + t^{2}}

I = \frac{l n 3}{4} - \frac{l n 2}{2} - {[s i n x]}_{0}^{π / 6} + 2 \int_{0}^{t a n (π / 12)} \frac{d t}{1 - t^{2}}

I = \frac{l n 3}{4} - \frac{l n 2}{2} - \frac{1}{2} + {[l n ∣ \frac{1 + t}{1 - t} ∣]}_{0}^{t a n (π / 12)}

t a n (π / 12) = \sqrt{\frac{2 - \sqrt{3}}{2 + \sqrt{3}}} = \sqrt{\frac{{(2 - \sqrt{3})}^{2}}{4 - 3}} = 2 - \sqrt{3}

I = \frac{l n 3}{4} - \frac{l n 2}{2} - \frac{1}{2} + l n ∣ \frac{1 + 2 - \sqrt{3}}{1 - 2 + \sqrt{3}} ∣

= \frac{l n 3}{4} - \frac{l n 2}{2} - \frac{1}{2} + l n (3 - \sqrt{3}) - l n (\sqrt{3} - 1)

Commented by Abdo msup. last updated on 17/Dec/18

t h a n k y o u s i r s m a i l .