$\int_{0}^{\frac{\pi}{2}} \cos(2nx) dx$ を計算せよ。解析学定積分三角関数積分2025/8/41. 問題の内容∫0π2cos(2nx)dx\int_{0}^{\frac{\pi}{2}} \cos(2nx) dx∫02πcos(2nx)dx を計算せよ。2. 解き方の手順まず、cos(2nx)\cos(2nx)cos(2nx) の不定積分を求めます。∫cos(2nx)dx=12nsin(2nx)+C\int \cos(2nx) dx = \frac{1}{2n} \sin(2nx) + C∫cos(2nx)dx=2n1sin(2nx)+C次に、定積分を計算します。∫0π2cos(2nx)dx=[12nsin(2nx)]0π2=12nsin(2n⋅π2)−12nsin(2n⋅0)=12nsin(nπ)−12nsin(0)\int_{0}^{\frac{\pi}{2}} \cos(2nx) dx = \left[ \frac{1}{2n} \sin(2nx) \right]_{0}^{\frac{\pi}{2}} = \frac{1}{2n} \sin(2n \cdot \frac{\pi}{2}) - \frac{1}{2n} \sin(2n \cdot 0) = \frac{1}{2n} \sin(n\pi) - \frac{1}{2n} \sin(0)∫02πcos(2nx)dx=[2n1sin(2nx)]02π=2n1sin(2n⋅2π)−2n1sin(2n⋅0)=2n1sin(nπ)−2n1sin(0)sin(nπ)=0\sin(n\pi) = 0sin(nπ)=0 および sin(0)=0\sin(0) = 0sin(0)=0 なので、12nsin(nπ)−12nsin(0)=12n⋅0−12n⋅0=0\frac{1}{2n} \sin(n\pi) - \frac{1}{2n} \sin(0) = \frac{1}{2n} \cdot 0 - \frac{1}{2n} \cdot 0 = 02n1sin(nπ)−2n1sin(0)=2n1⋅0−2n1⋅0=03. 最終的な答え0