integral of sin(\sqrt[4]{x-1})

解答

\int sin (4 x - 1) d x

4 (- cos (4 x - 1) (x - 1)^{\frac{3}{4}} + 3 (sin (4 x - 1) x - 1 - 2 (- cos (4 x - 1) 4 x - 1 + sin (4 x - 1)))) + C

求解步骤

\int sin (4 x - 1) d x

使用换元积分法

= \int 4 u^{3} sin (u) d u

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \int u^{3} sin (u) d u

使用分布积分法

= 4 (- u^{3} cos (u) - \int - 3 u^{2} cos (u) d u)

\int - 3 u^{2} cos (u) d u = - 3 (u^{2} sin (u) - 2 (- u cos (u) + sin (u)))

= 4 (- u^{3} cos (u) - (- 3 (u^{2} sin (u) - 2 (- u cos (u) + sin (u)))))

u = 4 x - 1

代回

= 4 (- (4 x - 1)^{3} cos (4 x - 1) - (- 3 ((4 x - 1)^{2} sin (4 x - 1) - 2 (- 4 x - 1 cos (4 x - 1) + sin (4 x - 1)))))

化简

4 (- (4 x - 1)^{3} cos (4 x - 1) - (- 3 ((4 x - 1)^{2} sin (4 x - 1) - 2 (- 4 x - 1 cos (4 x - 1) + sin (4 x - 1))))) : 4 (- cos (4 x - 1) (x - 1)^{\frac{3}{4}} + 3 (sin (4 x - 1) x - 1 - 2 (- cos (4 x - 1) 4 x - 1 + sin (4 x - 1))))

= 4 (- cos (4 x - 1) (x - 1)^{\frac{3}{4}} + 3 (sin (4 x - 1) x - 1 - 2 (- cos (4 x - 1) 4 x - 1 + sin (4 x - 1))))

解答补常数

= 4 (- cos (4 x - 1) (x - 1)^{\frac{3}{4}} + 3 (sin (4 x - 1) x - 1 - 2 (- cos (4 x - 1) 4 x - 1 + sin (4 x - 1)))) + C

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