https://spherical-harmonics.hateblo.jp/entry/Kyodai/2025/Rikei_1

2025.03.02記

[1] 問２次の定積分の値を求めよ．
(1) $\displaystyle\int_0^{\sqrt{3}}\dfrac{x\sqrt{x^2+1}+2x^3+1}{x^2+1}\, dx$
(2) $\displaystyle\int_0^\frac{\pi}{2}\sqrt{\dfrac{1-\cos x}{1+\cos x}}\, dx$

2025.03.02記
(2)の式変形は X で定期的に見る．

[解答]
(1) $\displaystyle\int_0^{\sqrt{3}}\dfrac{x\sqrt{x^2+1}+2x^3+1}{x^2+1}\, dx$ $=\displaystyle\int_0^{\sqrt{3}}\left(\dfrac{x}{\sqrt{x^2+1}}+2x-\dfrac{2x}{x^2+1}+\dfrac{1}{x^2+1}\right)\, dx$ $=\Bigl[\sqrt{x^2+1}+x^2-\log(x^2+1)\Bigr]_0^{\sqrt{3}}+ \Bigl[\theta\Bigr]_0^{\sqrt{\pi/3}}$ $=(2-1)+(3-0)-(\log 4-\log 1)+\left(\dfrac{\pi}{3}-0\right)$ $=4-2\log 2+\dfrac{\pi}{3}$
となる．

(2) $\displaystyle\int_0^\frac{\pi}{2}\sqrt{\dfrac{1-\cos x}{1+\cos x}}\, dx$ $=\displaystyle\int_0^\frac{\pi}{2}\left|\tan\dfrac{x}{2}\right|\, dx$ $=\displaystyle\int_0^\frac{\pi}{2}\tan\dfrac{x}{2}\, dx$ $=\Bigl[-2\log \left|\cos\dfrac{x}{2}\right|\Bigr]_0^{\frac{\pi}{2}}$ $=-2\log\dfrac{1}{\sqrt{2}}$ $=\log 2$
となる．