1935年(昭和10年)東京帝國大學工學部-數學[3] - [別館]球面倶楽部零八式markIISR

[3] 次ノ定積分ノ値ヲ求メヨ．
(a) $\displaystyle\int_{0}^{3}\dfrac{dx}{\sqrt{|\,x(x-2)\,|}}$
(b) $\displaystyle\int_{0}^{\pi}\dfrac{x\sin x}{1+\cos^2 x}dx$

2019.04.04記

[解答]
(a) $\displaystyle\int_{0}^{3}\dfrac{dx}{\sqrt{|\,x(x-2)\,|}}=\int_{0}^{2}\dfrac{dx}{\sqrt{x(2-x)}}+\int_{2}^{3}\dfrac{dx}{\sqrt{x(x-2)}}$ $=\displaystyle\int_{-1}^{1}\dfrac{dx}{\sqrt{1-x^2}}+\int_{1}^{2}\dfrac{dx}{\sqrt{x^2-1}}$ $=\Bigl[\sin^{-1}x\Bigr]_{-1}^{1}+\Bigl[\log|x+\sqrt{x^2-1}|\Bigr]_{1}^{2}$ $=\pi+\log(2+\sqrt{3})$

(b) $\displaystyle\int_{0}^{\pi}\dfrac{x\sin x}{1+\cos^2 x}dx$ $=\Bigl[-x\arctan\cos x\Bigr]_0^{\pi}+\displaystyle\int_{0}^{\pi}\arctan\cos xdx$ $=\Bigl[-x\arctan\cos x\Bigr]_0^{\pi}$ （ $\arctan\cos x$ は $x=\dfrac{\pi}{2}$ で点対称）
$=\dfrac{\pi^2}{4}$