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受欢迎的微积分 >

integral of sec(1-z)tan(1-z)

解答

\int sec (1 - z) tan (1 - z) d z

解答

- sec (1 - z) + C

求解步骤

\int sec (1 - z) tan (1 - z) d z

使用换元积分法

= \int - sec (u) tan (u) d u

提出常数:

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

= - \int sec (u) tan (u) d u

使用换元积分法

= - \int 1 d v

常数积分:

\int a d x = a x

= - 1 \cdot v

代回

= - 1 \cdot sec (1 - z)

化简

= - sec (1 - z)

解答补常数

= - sec (1 - z) + C

作图

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