次の2重積分を計算します。 $\int_{0}^{\pi}\int_{0}^{\pi-y} \sin(x+y) \, dx \, dy$解析学積分多重積分二重積分三角関数2025/6/121. 問題の内容次の2重積分を計算します。∫0π∫0π−ysin(x+y) dx dy\int_{0}^{\pi}\int_{0}^{\pi-y} \sin(x+y) \, dx \, dy∫0π∫0π−ysin(x+y)dxdy2. 解き方の手順まず、xxx に関する積分を行います。∫0π−ysin(x+y) dx=[−cos(x+y)]0π−y=−cos(π)+cos(y)=1+cos(y)\int_{0}^{\pi-y} \sin(x+y) \, dx = [-\cos(x+y)]_{0}^{\pi-y} = -\cos(\pi) + \cos(y) = 1 + \cos(y)∫0π−ysin(x+y)dx=[−cos(x+y)]0π−y=−cos(π)+cos(y)=1+cos(y)次に、yyy に関する積分を行います。∫0π(1+cos(y)) dy=[y+sin(y)]0π=(π+sin(π))−(0+sin(0))=π+0−0−0=π\int_{0}^{\pi} (1 + \cos(y)) \, dy = [y + \sin(y)]_{0}^{\pi} = (\pi + \sin(\pi)) - (0 + \sin(0)) = \pi + 0 - 0 - 0 = \pi∫0π(1+cos(y))dy=[y+sin(y)]0π=(π+sin(π))−(0+sin(0))=π+0−0−0=π3. 最終的な答えπ\piπ