定積分 $\int_{0}^{\pi} |\sin x - \sqrt{3} \cos x| dx$ の値を求める。解析学定積分三角関数絶対値積分2025/7/91. 問題の内容定積分 ∫0π∣sinx−3cosx∣dx\int_{0}^{\pi} |\sin x - \sqrt{3} \cos x| dx∫0π∣sinx−3cosx∣dx の値を求める。2. 解き方の手順まず、被積分関数 sinx−3cosx\sin x - \sqrt{3} \cos xsinx−3cosx を合成します。sinx−3cosx=2(12sinx−32cosx)=2(sinxcosπ3−cosxsinπ3)=2sin(x−π3)\sin x - \sqrt{3} \cos x = 2(\frac{1}{2}\sin x - \frac{\sqrt{3}}{2}\cos x) = 2(\sin x \cos \frac{\pi}{3} - \cos x \sin \frac{\pi}{3}) = 2\sin(x - \frac{\pi}{3})sinx−3cosx=2(21sinx−23cosx)=2(sinxcos3π−cosxsin3π)=2sin(x−3π).したがって、積分は∫0π∣2sin(x−π3)∣dx=2∫0π∣sin(x−π3)∣dx\int_{0}^{\pi} |2\sin(x - \frac{\pi}{3})| dx = 2\int_{0}^{\pi} |\sin(x - \frac{\pi}{3})| dx∫0π∣2sin(x−3π)∣dx=2∫0π∣sin(x−3π)∣dx.次に、∣sin(x−π3)∣|\sin(x - \frac{\pi}{3})|∣sin(x−3π)∣ の符号を調べます。sin(x−π3)=0\sin(x - \frac{\pi}{3}) = 0sin(x−3π)=0 となるのは、x−π3=0,πx - \frac{\pi}{3} = 0, \pix−3π=0,π より、x=π3,4π3x = \frac{\pi}{3}, \frac{4\pi}{3}x=3π,34π。積分区間 [0,π][0, \pi][0,π] において、sin(x−π3)\sin(x - \frac{\pi}{3})sin(x−3π) は 0≤x≤π30 \le x \le \frac{\pi}{3}0≤x≤3π で負、π3≤x≤π \frac{\pi}{3} \le x \le \pi3π≤x≤π で正です。したがって、2∫0π∣sin(x−π3)∣dx=2[−∫0π3sin(x−π3)dx+∫π3πsin(x−π3)dx]2\int_{0}^{\pi} |\sin(x - \frac{\pi}{3})| dx = 2\left[ -\int_{0}^{\frac{\pi}{3}} \sin(x - \frac{\pi}{3}) dx + \int_{\frac{\pi}{3}}^{\pi} \sin(x - \frac{\pi}{3}) dx \right]2∫0π∣sin(x−3π)∣dx=2[−∫03πsin(x−3π)dx+∫3ππsin(x−3π)dx]=2[cos(x−π3)∣0π3−cos(x−π3)∣π3π]= 2\left[ \cos(x - \frac{\pi}{3}) \Big|_{0}^{\frac{\pi}{3}} - \cos(x - \frac{\pi}{3}) \Big|_{\frac{\pi}{3}}^{\pi} \right]=2[cos(x−3π)03π−cos(x−3π)3ππ]=2[(cos0−cos(−π3))−(cos(2π3)−cos0)]= 2\left[ (\cos 0 - \cos(-\frac{\pi}{3})) - (\cos(\frac{2\pi}{3}) - \cos 0) \right]=2[(cos0−cos(−3π))−(cos(32π)−cos0)]=2[(1−12)−(−12−1)]= 2\left[ (1 - \frac{1}{2}) - (-\frac{1}{2} - 1) \right]=2[(1−21)−(−21−1)]=2[12−(−32)]=2[12+32]=2⋅42=4= 2\left[ \frac{1}{2} - (-\frac{3}{2}) \right] = 2\left[ \frac{1}{2} + \frac{3}{2} \right] = 2 \cdot \frac{4}{2} = 4=2[21−(−23)]=2[21+23]=2⋅24=4.3. 最終的な答え4