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受欢迎的微积分 >

integral of (-1)/4 sin(2x)

解答

\int \frac{- 1}{4} sin (2 x) d x

解答

\frac{1}{8} cos (2 x) + C

求解步骤

\int \frac{- 1}{4} sin (2 x) d x

提出常数:

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

= \frac{- 1}{4} \cdot \int sin (2 x) d x

化简

= (- \frac{1}{4}) \cdot \int sin (2 x) d x

使用换元积分法

= (- \frac{1}{4}) \cdot \int sin (u) \frac{1}{2} d u

提出常数:

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

= (- \frac{1}{4}) \frac{1}{2} \cdot \int sin (u) d u

使用积分常见定则:

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

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

u = 2 x

代回

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

化简

(- \frac{1}{4}) \frac{1}{2} (- cos (2 x)) : \frac{1}{8} cos (2 x)

= \frac{1}{8} cos (2 x)

解答补常数

= \frac{1}{8} cos (2 x) + C

作图

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