展開する (cos(3x+1)sec(2x+1))(12x(2x^3+1)^2)

解

展開する

(cos (3 x + 1) sec (2 x + 1)) (12 x (2 x^{3} + 1)^{2})

48 x^{7} cos (3 x + 1) sec (2 x + 1) + 48 x^{4} cos (3 x + 1) sec (2 x + 1) + 12 x cos (3 x + 1) sec (2 x + 1)

解答ステップ

(cos (3 x + 1) sec (2 x + 1)) (12 x (2 x^{3} + 1)^{2})

括弧を削除する:

(a) = a

= cos (3 x + 1) sec (2 x + 1) \cdot 12 x (2 x^{3} + 1)^{2}

(2 x^{3} + 1)^{2} = 4 x^{6} + 4 x^{3} + 1

= 12 x cos (3 x + 1) sec (2 x + 1) (4 x^{6} + 4 x^{3} + 1)

= 12 x cos (3 x + 1) sec (2 x + 1) (4 x^{6} + 4 x^{3} + 1)

括弧を分配する

= cos (3 x + 1) sec (2 x + 1) \cdot 12 x \cdot 4 x^{6} + cos (3 x + 1) sec (2 x + 1) \cdot 12 x \cdot 4 x^{3} + cos (3 x + 1) sec (2 x + 1) \cdot 12 x \cdot 1

= 12 \cdot 4 x^{6} x cos (3 x + 1) sec (2 x + 1) + 12 \cdot 4 x^{3} x cos (3 x + 1) sec (2 x + 1) + 12 \cdot 1 \cdot x cos (3 x + 1) sec (2 x + 1)

簡素化

12 \cdot 4 x^{6} x cos (3 x + 1) sec (2 x + 1) + 12 \cdot 4 x^{3} x cos (3 x + 1) sec (2 x + 1) + 12 \cdot 1 \cdot x cos (3 x + 1) sec (2 x + 1) : 48 x^{7} cos (3 x + 1) sec (2 x + 1) + 48 x^{4} cos (3 x + 1) sec (2 x + 1) + 12 x cos (3 x + 1) sec (2 x + 1)

= 48 x^{7} cos (3 x + 1) sec (2 x + 1) + 48 x^{4} cos (3 x + 1) sec (2 x + 1) + 12 x cos (3 x + 1) sec (2 x + 1)