integral of tcos(2t)

解答

\int t cos (2 t) d t

\frac{1}{4} (2 t sin (2 t) + cos (2 t)) + C

求解步骤

\int t cos (2 t) d t

使用换元积分法

= \int \frac{u cos ( u )}{4} d u

提出常数:

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

= \frac{1}{4} \cdot \int u cos (u) d u

使用分布积分法

= \frac{1}{4} (u sin (u) - \int sin (u) d u)

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

= \frac{1}{4} (u sin (u) - (- cos (u)))

u = 2 t

代回

= \frac{1}{4} (2 t sin (2 t) - (- cos (2 t)))

化简

= \frac{1}{4} (2 t sin (2 t) + cos (2 t))

解答补常数

= \frac{1}{4} (2 t sin (2 t) + cos (2 t)) + C

\int \frac{x ( x ^{2} - 1 )}{x ^{2} + 1} d x

(x + y - 3) d x + (x + y + 4) d y = 0

n = 0 \sum \infty - 1^{n + 1}

\frac{d y}{d x} = \frac{x + 2}{e ^{y - 1}}

\frac{d}{d x} (4 (e^{x} cos (x) - e^{x} sin (x)))