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
$$ \displaystyle \int_0^\frac{\pi}{2} \frac{\arctan{\frac{\tan{x}}{\cos{x}}}}{\tan{x}}dx $$
[解説]
$
\begin{eqnarray*}
&&\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{x}dx\\
&=&\int_0^1 \frac{\arctan{\frac{t}{1-t^2}}}{t}dt\hspace{15pt}(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\hspace{15pt}(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}^\infty \frac{(-1)^k}{2k+1}t^{2k+1}dt\\
&=&\frac{4}{3} \sum_{k=0}^\infty \frac{(-1)^k}{2k+1} \int_0^1 t^{2k}dt\\
&=&\frac{4}{3} \sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)^2}\\
&=&\frac{4}{3}G
\end{eqnarray*}
$
よって、この問題の解答は$\displaystyle \frac{4}{3}G$となります。