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

..... nice .... .... math.... prove that: 𝛗=∫_0 ^( (π/2)) ((ln(1+sin(x).cos(x)))/(tan(x)))dx=((5π^2 )/(72))

$$\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:…..\:{nice}\:….\:….\:{math}…. \\ $$$$\:\:\:{prove}\:{that}: \\ $$$$\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \frac{{ln}\left(\mathrm{1}+{sin}\left({x}\right).{cos}\left({x}\right)\right)}{{tan}\left({x}\right)}{dx}=\frac{\mathrm{5}\pi^{\mathrm{2}} }{\mathrm{72}} \\ $$

Commented by liki last updated on 24/Apr/21

Commented by liki last updated on 24/Apr/21

$$..\mathrm{help}\:\mathrm{pls} \\ $$

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

$${to}\:{liki}: \\ $$$${please}\:{don}'{t}\:{insert}\:{your}\:{question}\:{into} \\ $$$${the}\:{threads}\:{of}\:{other}\:{people}\:{for}\:{other} \\ $$$${questions}!\:{just}\:{open}\:{a}\:{new}\:{thread}\:{for} \\ $$$${your}\:{own}\:{question}!\:{if}\:{you}\:{don}'{t}\:{know} \\ $$$${how},\:{just}\:{read}\:{the}\:{ForumHelp}! \\ $$

Answered by qaz last updated on 24/Apr/21

φ=∫_0 ^(π/2) ((ln(1+sin xcos x))/(tan x))dx =∫_0 ^(π/2) ((ln(1+((tan x)/(1+tan^2 x))))/(tan x))dx =∫_0 ^(π/2) ((ln(1+tan x+tan^2 x)−ln(1+tan^2 x))/(tan x))dx =∫_0 ^∞ ((ln(1+x+x^2 )−ln(1+x^2 ))/(x(1+x^2 )))dx =∫_0 ^1 ((ln(1+x+x^2 )−ln(1+x^2 ))/(x(1+x^2 )))dx+∫_1 ^∞ ((ln(1+x+x^2 )−ln(1+x^2 ))/((1/x)(1+(1/x^2 ))))(−(1/x^2 ))dx =∫_0 ^1 ((ln(1+x+x^2 )−ln(1+x^2 ))/(x(1+x^2 )))dx+∫_0 ^1 ((ln(1+x+x^2 )−ln(1+x^2 ))/(1+x^2 ))xdx =∫_0 ^1 ((ln(1+x+x^2 )−ln(1+x^2 ))/x)dx =∫_0 ^1 ((ln(1−x^3 )−ln(1−x))/x)dx−(1/2)η(2) =−Σ_(n=1) ^∞ ∫_0 ^1 (x^(3n−1) /n)dx+(π^2 /6)−(π^2 /(24)) =−Σ_(n=1) ^∞ (1/(3n^2 ))+(π^2 /8) =−(π^2 /(18))+(π^2 /8) =((5π^2 )/(72))

$$\phi=\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \frac{{ln}\left(\mathrm{1}+\mathrm{sin}\:{x}\mathrm{cos}\:{x}\right)}{\mathrm{tan}\:{x}}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \frac{{ln}\left(\mathrm{1}+\frac{\mathrm{tan}\:{x}}{\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} {x}}\right)}{\mathrm{tan}\:{x}}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \frac{{ln}\left(\mathrm{1}+\mathrm{tan}\:{x}+\mathrm{tan}\:^{\mathrm{2}} {x}\right)−{ln}\left(\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} {x}\right)}{\mathrm{tan}\:{x}}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \frac{{ln}\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} \right)−{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{{x}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} \right)−{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{{x}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{dx}+\int_{\mathrm{1}} ^{\infty} \frac{{ln}\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} \right)−{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{\frac{\mathrm{1}}{{x}}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)}\left(−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right){dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} \right)−{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{{x}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{dx}+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} \right)−{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{\mathrm{1}+{x}^{\mathrm{2}} }{xdx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} \right)−{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{{x}}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}−{x}^{\mathrm{3}} \right)−{ln}\left(\mathrm{1}−{x}\right)}{{x}}{dx}−\frac{\mathrm{1}}{\mathrm{2}}\eta\left(\mathrm{2}\right) \\ $$$$=−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{3}{n}−\mathrm{1}} }{{n}}{dx}+\frac{\pi^{\mathrm{2}} }{\mathrm{6}}−\frac{\pi^{\mathrm{2}} }{\mathrm{24}} \\ $$$$=−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{3}{n}^{\mathrm{2}} }+\frac{\pi^{\mathrm{2}} }{\mathrm{8}} \\ $$$$=−\frac{\pi^{\mathrm{2}} }{\mathrm{18}}+\frac{\pi^{\mathrm{2}} }{\mathrm{8}} \\ $$$$=\frac{\mathrm{5}\pi^{\mathrm{2}} }{\mathrm{72}} \\ $$

Commented by mnjuly1970 last updated on 24/Apr/21

$$\:\:{great},\:{thanks}\:{alot}\:{master}… \\ $$

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