integral from 2 to 1/x of tan^2(t)

解答

\int_{2}^{\frac{1}{x}} tan^{2} (t) d t

- \frac{1}{x} + 2 + tan (\frac{1}{x}) - tan (2)

求解步骤

\int_{2}^{\frac{1}{x}} tan^{2} (t) d t

使用三角恒等式改写

= \int_{2}^{\frac{1}{x}} - 1 + sec^{2} (t) d t

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int_{2}^{\frac{1}{x}} 1 d t + \int_{2}^{\frac{1}{x}} sec^{2} (t) d t

\int_{2}^{\frac{1}{x}} 1 d t = \frac{1}{x} - 2

\int_{2}^{\frac{1}{x}} sec^{2} (t) d t = tan (\frac{1}{x}) - tan (2)

= - (\frac{1}{x} - 2) + tan (\frac{1}{x}) - tan (2)

化简

= - \frac{1}{x} + 2 + tan (\frac{1}{x}) - tan (2)

t an g e n t f (x) = x \cdot cot (x), at x = \frac{π}{4}

d er i v a t i v e f (x) = \frac{x + x}{x ^{3}}

\frac{d}{d x} (\frac{y}{y + x ^{2}})

\frac{d}{d x} (e^{x^{2}} sin (x))

\frac{d x}{d t} = 2 x (10 - x)