$\int \sin(5\theta)\cos(5\theta) d\theta$ を計算する問題です。解析学積分三角関数積和の公式不定積分2025/5/151. 問題の内容∫sin(5θ)cos(5θ)dθ\int \sin(5\theta)\cos(5\theta) d\theta∫sin(5θ)cos(5θ)dθ を計算する問題です。2. 解き方の手順三角関数の積の積分なので、三角関数の積和の公式を利用して計算します。sin(A)cos(B)=12[sin(A+B)+sin(A−B)]\sin(A)\cos(B) = \frac{1}{2}[\sin(A+B) + \sin(A-B)]sin(A)cos(B)=21[sin(A+B)+sin(A−B)]この公式を適用すると、sin(5θ)cos(5θ)=12[sin(5θ+5θ)+sin(5θ−5θ)]=12[sin(10θ)+sin(0)]=12sin(10θ)\sin(5\theta)\cos(5\theta) = \frac{1}{2}[\sin(5\theta+5\theta) + \sin(5\theta-5\theta)] = \frac{1}{2}[\sin(10\theta) + \sin(0)] = \frac{1}{2}\sin(10\theta)sin(5θ)cos(5θ)=21[sin(5θ+5θ)+sin(5θ−5θ)]=21[sin(10θ)+sin(0)]=21sin(10θ)したがって、∫sin(5θ)cos(5θ)dθ=∫12sin(10θ)dθ\int \sin(5\theta)\cos(5\theta) d\theta = \int \frac{1}{2}\sin(10\theta) d\theta∫sin(5θ)cos(5θ)dθ=∫21sin(10θ)dθ∫12sin(10θ)dθ=12∫sin(10θ)dθ\int \frac{1}{2}\sin(10\theta) d\theta = \frac{1}{2} \int \sin(10\theta) d\theta∫21sin(10θ)dθ=21∫sin(10θ)dθここで、∫sin(ax)dx=−1acos(ax)+C\int \sin(ax) dx = -\frac{1}{a}\cos(ax) + C∫sin(ax)dx=−a1cos(ax)+C を利用すると、12∫sin(10θ)dθ=12(−110cos(10θ))+C=−120cos(10θ)+C\frac{1}{2} \int \sin(10\theta) d\theta = \frac{1}{2} \left( -\frac{1}{10}\cos(10\theta) \right) + C = -\frac{1}{20}\cos(10\theta) + C21∫sin(10θ)dθ=21(−101cos(10θ))+C=−201cos(10θ)+C3. 最終的な答え−120cos(10θ)+C-\frac{1}{20}\cos(10\theta) + C−201cos(10θ)+C