積分作ってみた

$I_2={\displaystyle {\int }_0^{\frac{\pi }{2}}{\displaystyle \frac{1-\sin x+x\cos x}{x(1+\sin x)+2\cos x}}\mathrm{d}x}$
$I_3={\displaystyle {\int }_0^{\frac{\pi }{2}}{\displaystyle \frac{1-\sin x}{x(1+\sin x)+2\cos x}}\mathrm{d}x}$
を考える。

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

$I_3={\displaystyle {\int }_{-\frac{\pi }{4}}^0{\displaystyle \frac{1+\sin 2t}{-t(1-\sin 2t)+\cos 2t}}\mathrm{d}t}$
$={\displaystyle {\int }_{-\frac{\pi }{4}}^0{\displaystyle \frac{(\cos t+\sin t{)}^2}{-t(\cos t-\sin t{)}^2+({\cos }^{2}t-{\sin }^{2}t)}}\mathrm{d}t}$

分母・分子を$(\cos t-\sin t{)}^2$で割る

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

$={\displaystyle {\int }_{-\frac{\pi }{4}}^0{\displaystyle \frac{{\tan }^2(t+{\displaystyle \frac{\pi }{4}})}{-t+\tan (t+{\displaystyle \frac{\pi }{4}})}}\mathrm{d}t}$

$t+{\displaystyle \frac{\pi }{4}}=u$と置換(この置換はしなくてもよい)

$I_3={\displaystyle {\int }_0^{\frac{\pi }{4}}{\displaystyle \frac{{\tan }^{2}u}{\tan u-u+{\displaystyle \frac{\pi }{4}}}}\mathrm{d}u}$
$=[\log (\tan u-u+{\displaystyle \frac{\pi }{4}}){]}_0^{\frac{\pi }{4}}$
$=-\log {\displaystyle \frac{\pi }{4}}$

$I=I_2-I_3$
$=\log {\displaystyle \frac{\pi }{2}}+\log {\displaystyle \frac{\pi }{4}}$
$=\log {\displaystyle \frac{{\pi }^2}{8}}$

$I$について、$x=2t$と置換

$I=2{\displaystyle {\int }_0^{\frac{\pi }{4}}{\displaystyle \frac{t({\cos }^{2}t-{\sin }^{2}t)}{t(\cos t+\sin t{)}^2+({\cos }^{2}t-{\sin }^{2}t)}}}$
$=2{\displaystyle {\int }_0^{\frac{\pi }{4}}{\displaystyle \frac{t(\cos t-\sin t)}{t(\cos t+\sin t)+(\cos t-\sin t)}}}$
$=2[\log (t(\cos t+\sin t)+(\cos t-\sin t)){]}_0^{\frac{\pi }{4}}$
$=2\log {\displaystyle \frac{\sqrt{2}\pi }{4}}$
$=\log {\displaystyle \frac{{\pi }^2}{8}}$

(えっ$\dots$)