integral of e^{tan(9x)}sec^2(9x)

解答

\int e^{t a n (9 x)} sec^{2} (9 x) d x

\frac{1}{9} e^{t a n (9 x)} + C

求解步骤

\int e^{t a n (9 x)} sec^{2} (9 x) d x

使用换元积分法

= \int e^{t a n (u)} sec^{2} (u) \frac{1}{9} d u

提出常数:

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

= \frac{1}{9} \cdot \int e^{t a n (u)} sec^{2} (u) d u

使用换元积分法

= \frac{1}{9} \cdot \int e^{v} d v

使用积分常见定则:

\int e^{v} d v = e^{v}

= \frac{1}{9} e^{v}

代回

= \frac{1}{9} e^{t a n (9 x)}

解答补常数

= \frac{1}{9} e^{t a n (9 x)} + C

t an g e n t f (x) = x^{2} + 15 x + 56, at x = 2

a re a 6 x, x ((2 1^{2} - x^{2}))

t an g e n t y = e^{x} sin (x), (0, 0)

\int x^{- 1} + x^{\frac{1}{3}} + \frac{5}{3} d x

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