https://spherical-harmonics.hateblo.jp/entry/Todai/1929/Kougaku_3

2022.09.01記

[3] $\displaystyle\int\dfrac{\cos\theta(p+\sin\theta)}{\cos^2\theta-p\sin\theta}d\theta$ ヲ計算セヨ．

2022.09.07記

[解答]
$t=\sin\theta$ とおくと $dt=\cos\theta\,d\theta$ により
$I=\displaystyle\int\dfrac{\cos\theta(p+\sin\theta)}{\cos^2\theta-p\sin\theta}d\theta$
$=\displaystyle\int\dfrac{p+t}{1-t^2-pt}\, dt$
$=\displaystyle\int\dfrac{p/2+t}{1-t^2-pt}\, dt$ $+\displaystyle\int\dfrac{p/2}{1-t^2-pt}\, dt$
$=-\dfrac{1}{2}\log |1-t^2-pt|$ $+\dfrac{p}{2}\displaystyle\int\dfrac{1}{1+p^2/4-(t+p/2)^2}\, dt$
である．ここで
$\dfrac{1}{(t-\alpha)(t-\beta)}=\dfrac{1}{\beta-\alpha}\left(\dfrac{1}{t-\beta}-\dfrac{1}{t-\alpha}\right)$
に注意すると
$I=-\dfrac{1}{2}\log |1-t^2-pt|$ $-\dfrac{p}{2\sqrt{p^2+4}}\log\left| \dfrac{2t-p-\sqrt{p^2+4}}{2t-p+\sqrt{p^2+4}} \right| +(積分定数)$
$=-\dfrac{1}{2}\log |\cos^2\theta-p\sin\theta|$ $-\dfrac{p}{2\sqrt{p^2+4}}\log\left| \dfrac{2\sin\theta-p-\sqrt{p^2+4}}{2\sin\theta-p+\sqrt{p^2+4}} \right| +(積分定数)$