積分解説06

2020/10/10に積分コンテストで出題した問題です。
https://twitter.com/sounansya_29/status/1314883373463031808?s=21

[解説]
$\begin{array}{rcl}& & {\int }_0^{\frac{\pi }{2}}\frac{\arctan \frac{\tan x}{\cos x}}{\tan x}dx\\ & =& {\int }_0^{\frac{\pi }{2}}\frac{\arctan \frac{\sin x}{1-{\sin }^{2}x}}{\sin x}\cos xdx\\ & =& {\int }_0^1\frac{\arctan \frac{t}{1-t^2}}{t}dt(t=\sin x)\\ & =& {\int }_0^1\frac{\arctan \frac{t+t^3}{1-t^4}}{t}dt\\ & =& {\int }_0^1\frac{\arctan t+\arctan t^3}{t}dt\\ & =& {\int }_0^1\frac{\arctan t}{t}dt+\frac{1}{3}{\int }_0^1\frac{\arctan u}{\sqrt[3]{u}}{u}^{-\frac{2}{3}}du(t=\sqrt[3]{u})\\ & =& \frac{4}{3}{\int }_0^1\frac{\arctan t}{t}dt\\ & =& \frac{4}{3}{\int }_0^1\frac{1}{t}\sum _{k=0}^{\mathrm{\infty }}\frac{(-1{)}^k}{2k+1}{t}^{2k+1}dt\\ & =& \frac{4}{3}\sum _{k=0}^{\mathrm{\infty }}\frac{(-1{)}^k}{2k+1}{\int }_0^1{t}^{2k}dt\\ & =& \frac{4}{3}\sum _{k=0}^{\mathrm{\infty }}\frac{(-1{)}^k}{(2k+1{)}^2}\\ & =& \frac{4}{3}G\end{array}$
よって、この問題の解答は${\displaystyle \frac{4}{3}G}$となります。