nice-math-prove-that-0-pi-2-ln-1-sin-x-cos-x-tan-x-dx-5pi-2-72-

Question Number 139217 by mnjuly1970 last updated on 24/Apr/21

\dots . . n i c e \dots . \dots . m a t h \dots .

p r o v e t h a t :

ϕ = \int_{0}^{\frac{π}{2}} \frac{l n (1 + s i n (x) . c o s (x))}{t a n (x)} d x = \frac{5 π^{2}}{72}

Commented by liki last updated on 24/Apr/21

Commented by liki last updated on 24/Apr/21

. . help pls

Commented by mr W last updated on 24/Apr/21

t o l i k i :

p l e a s e {d o n}^{'} t i n s e r t y o u r q u e s t i o n i n t o

t h e t h r e a d s o f o t h e r p e o p l e f o r o t h e r

q u e s t i o n s! j u s t o p e n a n e w t h r e a d f o r

y o u r o w n q u e s t i o n! i f y o u {d o n}^{'} t k n o w

h o w, j u s t r e a d t h e F o r u m H e l p!

Answered by qaz last updated on 24/Apr/21

ϕ = \int_{0}^{π / 2} \frac{l n (1 + \sin x \cos x)}{\tan x} d x

= \int_{0}^{π / 2} \frac{l n (1 + \frac{\tan x}{1 + \tan^{2} x})}{\tan x} d x

= \int_{0}^{π / 2} \frac{l n (1 + \tan x + \tan^{2} x) - l n (1 + \tan^{2} x)}{\tan x} d x

= \int_{0}^{\infty} \frac{l n (1 + x + x^{2}) - l n (1 + x^{2})}{x (1 + x^{2})} d x

= \int_{0}^{1} \frac{l n (1 + x + x^{2}) - l n (1 + x^{2})}{x (1 + x^{2})} d x + \int_{1}^{\infty} \frac{l n (1 + x + x^{2}) - l n (1 + x^{2})}{\frac{1}{x} (1 + \frac{1}{x^{2}})} (- \frac{1}{x^{2}}) d x

= \int_{0}^{1} \frac{l n (1 + x + x^{2}) - l n (1 + x^{2})}{x (1 + x^{2})} d x + \int_{0}^{1} \frac{l n (1 + x + x^{2}) - l n (1 + x^{2})}{1 + x^{2}} x d x

= \int_{0}^{1} \frac{l n (1 + x + x^{2}) - l n (1 + x^{2})}{x} d x

= \int_{0}^{1} \frac{l n (1 - x^{3}) - l n (1 - x)}{x} d x - \frac{1}{2} η (2)

= - \underset{n = 1}{\sum^{\infty}} \int_{0}^{1} \frac{x^{3 n - 1}}{n} d x + \frac{π^{2}}{6} - \frac{π^{2}}{24}

= - \underset{n = 1}{\sum^{\infty}} \frac{1}{3 n^{2}} + \frac{π^{2}}{8}

= - \frac{π^{2}}{18} + \frac{π^{2}}{8}

= \frac{5 π^{2}}{72}

Commented by mnjuly1970 last updated on 24/Apr/21

g r e a t, t h a n k s a l o t m a s t e r \dots