問題文全文(内容文)：
(1) $\displaystyle \int_0^1\sqrt{e^{1-t}}dt$
(2) $\displaystyle \int_0^{\frac{\pi}{2}}\dfrac{\cos2\theta}{\sin\theta+\cos\theta}d\theta$
(3) $\displaystyle \int_0^\pi\sin^4x\,dx$
(4) $\displaystyle \int_1^2\dfrac{\sqrt{x^2-4x+4}}{x}dx$
(1) $\displaystyle \int_0^1\sqrt{e^{1-t}}dt$
(2) $\displaystyle \int_0^{\frac{\pi}{2}}\dfrac{\cos2\theta}{\sin\theta+\cos\theta}d\theta$
(3) $\displaystyle \int_0^\pi\sin^4x\,dx$
(4) $\displaystyle \int_1^2\dfrac{\sqrt{x^2-4x+4}}{x}dx$
単元：
#積分とその応用#不定積分#数学(高校生)#数Ⅲ
指導講師：
理数個別チャンネル
問題文全文(内容文)：
(1) $\displaystyle \int_0^1\sqrt{e^{1-t}}dt$
(2) $\displaystyle \int_0^{\frac{\pi}{2}}\dfrac{\cos2\theta}{\sin\theta+\cos\theta}d\theta$
(3) $\displaystyle \int_0^\pi\sin^4x\,dx$
(4) $\displaystyle \int_1^2\dfrac{\sqrt{x^2-4x+4}}{x}dx$
(1) $\displaystyle \int_0^1\sqrt{e^{1-t}}dt$
(2) $\displaystyle \int_0^{\frac{\pi}{2}}\dfrac{\cos2\theta}{\sin\theta+\cos\theta}d\theta$
(3) $\displaystyle \int_0^\pi\sin^4x\,dx$
(4) $\displaystyle \int_1^2\dfrac{\sqrt{x^2-4x+4}}{x}dx$
投稿日：2026.08.16
