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受欢迎的微积分 >

integral of 1/6 sec(1/9 x)tan(1/9 x)

解答

\int \frac{1}{6} sec (\frac{1}{9} x) tan (\frac{1}{9} x) d x

解答

\frac{3}{2} sec (\frac{x}{9}) + C

求解步骤

\int \frac{1}{6} sec (\frac{1}{9} x) tan (\frac{1}{9} x) d x

提出常数:

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

= \frac{1}{6} \cdot \int sec (\frac{1}{9} x) tan (\frac{1}{9} x) d x

化简

sec (\frac{1}{9} x) tan (\frac{1}{9} x) : sec (\frac{x}{9}) tan (\frac{x}{9})

= \frac{1}{6} \cdot \int sec (\frac{x}{9}) tan (\frac{x}{9}) d x

使用换元积分法

= \frac{1}{6} \cdot \int sec (u) tan (u) \cdot 9 d u

提出常数:

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

= \frac{1}{6} \cdot 9 \cdot \int sec (u) tan (u) d u

使用换元积分法

= \frac{1}{6} \cdot 9 \cdot \int 1 d v

常数积分:

\int a d x = a x

= \frac{1}{6} \cdot 9 \cdot 1 \cdot v

代回

= \frac{1}{6} \cdot 9 \cdot 1 \cdot sec (\frac{x}{9})

化简

\frac{1}{6} \cdot 9 \cdot 1 \cdot sec (\frac{x}{9}) : \frac{3}{2} sec (\frac{x}{9})

= \frac{3}{2} sec (\frac{x}{9})

解答补常数

= \frac{3}{2} sec (\frac{x}{9}) + C

作图

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