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受欢迎的微积分 >

integral of tan(6x+1)

解答

\int tan (6 x + 1) d x

解答

- \frac{1}{6} ln ∣ cos (6 x + 1) ∣ + C

求解步骤

\int tan (6 x + 1) d x

使用换元积分法

= \int tan (u) \frac{1}{6} d u

提出常数:

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

= \frac{1}{6} \cdot \int tan (u) d u

使用三角恒等式改写

= \frac{1}{6} \cdot \int \frac{sin ( u )}{cos ( u )} d u

使用换元积分法

= \frac{1}{6} \cdot \int - \frac{1}{v} d v

提出常数:

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

= \frac{1}{6} (- \int \frac{1}{v} d v)

使用积分常见定则:

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{1}{6} (- ln ∣ v ∣)

代回

= \frac{1}{6} (- ln ∣ cos (6 x + 1) ∣)

化简

= - \frac{1}{6} ln ∣ cos (6 x + 1) ∣

解答补常数

= - \frac{1}{6} ln ∣ cos (6 x + 1) ∣ + C

作图

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