integral of sin(3t-3x)

解答

\int sin (3 t - 3 x) d x

\frac{1}{3} cos (3 t - 3 x) + C

求解步骤

\int sin (3 t - 3 x) d x

使用换元积分法

= \int - \frac{1}{3} sin (u) d u

提出常数:

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

= - \frac{1}{3} \cdot \int sin (u) d u

使用积分常见定则:

\int sin (u) d u = - cos (u)

= - \frac{1}{3} (- cos (u))

u = 3 t - 3 x

代回

= - \frac{1}{3} (- cos (3 t - 3 x))

化简

= \frac{1}{3} cos (3 t - 3 x)

解答补常数

= \frac{1}{3} cos (3 t - 3 x) + C

\int (x^{3} e^{x^{4}}) d x

t^{2} \frac{d y}{d t} + y = 0

d er i v a t i v e y = (1 + 5 x)^{10}

\frac{d}{d x} (- x + \frac{1}{x} + 1)

n = 1 \sum \infty (sin (1))^{n}