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

$$$$$$.....nice........math....$$$$provethat:$$$$\mathit{\varphi }={\int }_0^{\frac{\pi }{2}}\frac{ln(1+sin\left(x\right).cos\left(x\right))}{tan\left(x\right)}dx=\frac{5{\pi }^2}{72}$$

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

$$toliki:$$$$please{don}^{\prime }tinsertyourquestioninto$$$$thethreadsofotherpeopleforother$$$$questions!justopenanewthreadfor$$$$yourownquestion!ifyou{don}^{\prime }tknow$$$$how,justreadtheForumHelp!$$

Answered by qaz last updated on 24/Apr/21

$$\varphi ={\int }_0^{\pi /2}\frac{ln(1+\sin x\cos x)}{\tan x}dx$$$$={\int }_0^{\pi /2}\frac{ln(1+\frac{\tan x}{1+\tan {}^{2}x})}{\tan x}dx$$$$={\int }_0^{\pi /2}\frac{ln(1+\tan x+\tan {}^{2}x)-ln(1+\tan {}^{2}x)}{\tan x}dx$$$$={\int }_0^{\mathrm{\infty }}\frac{ln(1+x+x^2)-ln(1+x^2)}{x(1+x^2)}dx$$$$={\int }_0^1\frac{ln(1+x+x^2)-ln(1+x^2)}{x(1+x^2)}dx+{\int }_1^{\mathrm{\infty }}\frac{ln(1+x+x^2)-ln(1+x^2)}{\frac{1}{x}(1+\frac{1}{x^2})}(-\frac{1}{x^2})dx$$$$={\int }_0^1\frac{ln(1+x+x^2)-ln(1+x^2)}{x(1+x^2)}dx+{\int }_0^1\frac{ln(1+x+x^2)-ln(1+x^2)}{1+x^2}xdx$$$$={\int }_0^1\frac{ln(1+x+x^2)-ln(1+x^2)}{x}dx$$$$={\int }_0^1\frac{ln(1-x^3)-ln(1-x)}{x}dx-\frac{1}{2}\eta \left(2\right)$$$$=-\underset{n=1}{\sum ^{\mathrm{\infty }}}{\int }_0^1\frac{x^{3n-1}}{n}dx+\frac{{\pi }^2}{6}-\frac{{\pi }^2}{24}$$$$=-\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{3n^2}+\frac{{\pi }^2}{8}$$$$=-\frac{{\pi }^2}{18}+\frac{{\pi }^2}{8}$$$$=\frac{5{\pi }^2}{72}$$

Commented by mnjuly1970 last updated on 24/Apr/21

$$great,thanksalotmaster...$$

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