積分作ってみた

$I_2=\displaystyle\int_0^{\frac{\pi}{2}}\dfrac{1-\sin x+x\cos x}{x(1+\sin x)+2\cos x}\mathrm{d}x$
$I_3=\displaystyle\int_0^{\frac{\pi}{2}}\dfrac{1-\sin x}{x(1+\sin x)+2\cos x}\mathrm{d}x$
を考える。

$I_2=\biggl[ \log(x(1+\sin x)+2\cos x) \biggl]_0^{\frac{\pi}{2}}=\log \dfrac{\pi}{2}$
$I_3$について、$x=-2t$と置換

$I_3=\displaystyle\int_{-\frac{\pi}{4}}^0\dfrac{1+\sin 2t}{-t(1-\sin 2t)+\cos 2t}\mathrm{d}t$
$\hspace{12pt}=\displaystyle\int_{-\frac{\pi}{4}}^{0}\dfrac{(\cos t+\sin t)^2}{-t(\cos t-\sin t)^2+(\cos^2 t-\sin^2 t)}\mathrm{d}t$

分母・分子を$(\cos t-\sin t)^2$で割る

$I_3=\displaystyle\int_{-\frac{\pi}{4}}^0\dfrac{\Bigl(\dfrac{\cos t+\sin t}{\cos t-\sin t}\Bigl)^2}{-t+\Bigl(\dfrac{\cos t+\sin t}{\cos t-\sin t}\Bigl)}\mathrm{d}t$

$\hspace{12pt}=\displaystyle\int_{-\frac{\pi}{4}}^{0}\dfrac{\tan^2\Bigl(t+\dfrac{\pi}{4}\Bigl)}{-t+\tan\Bigl(t+\dfrac{\pi}{4}\Bigl)}\mathrm{d}t$

$t+\dfrac{\pi}{4}=u$と置換(この置換はしなくてもよい)

$I_3=\displaystyle\int_0^{\frac{\pi}{4}} \dfrac{\tan^2 u}{\tan u-u+\dfrac{\pi}{4}}\mathrm{d}u$
$\hspace{12pt}=\biggl[\log \Bigl(\tan u-u+\dfrac{\pi}{4}\Bigl)\biggl]_{0}^{\frac{\pi}{4}}$
$\hspace{12pt}=-\log \dfrac{\pi}{4}$

$I=I_2-I_3$
$\hspace{8.5pt}=\log \dfrac{\pi}{2}+\log \dfrac{\pi}{4}$
$\hspace{8.5pt}=\log\dfrac{\pi^2}{8}$

$I$について、$x=2t$と置換

$I=2\displaystyle\int_0^{\frac{\pi}{4}} \dfrac{t(\cos^2 t-\sin^2 t)}{t(\cos t+\sin t)^2 +(\cos^2t-\sin^2 t)}$
$\hspace{8.5pt}=2\displaystyle\int_0^{\frac{\pi}{4}}\dfrac{t(\cos t-\sin t)}{t(\cos t+\sin t) +(\cos t-\sin t)}$
$\hspace{8.5pt}=2\biggl[\log \Bigl(t(\cos t+\sin t)+(\cos t-\sin t)\Bigl)\biggl]_0^{\frac{\pi}{4}}$
$\hspace{8.5pt}=2\log \dfrac{\sqrt2\pi}{4}$
$\hspace{8.5pt}=\log\dfrac{\pi^2}{8}$

(えっ$\dots$)