integral of (sec(θ)tan(θ))/(sec^2(θ)-sec(θ))

解答

\int \frac{sec ( θ ) tan ( θ )}{sec ^{2} ( θ ) - sec ( θ )} d θ

- ln ∣ sec (θ) ∣ + ln ∣ sec (θ) - 1 ∣ + C

求解步骤

\int \frac{sec ( θ ) tan ( θ )}{sec ^{2} ( θ ) - sec ( θ )} d θ

使用换元积分法

= \int \frac{1}{u ^{2} - u} d u

将

\frac{1}{u ^{2} - u}

用部份分式展开:

- \frac{1}{u} + \frac{1}{u - 1}

= \int - \frac{1}{u} + \frac{1}{u - 1} d u

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 \frac{1}{u} d u + \int \frac{1}{u - 1} d u

\int \frac{1}{u} d u = ln ∣ u ∣

\int \frac{1}{u - 1} d u = ln ∣ u - 1 ∣

= - ln ∣ u ∣ + ln ∣ u - 1 ∣

u = sec (θ)

代回

= - ln ∣ sec (θ) ∣ + ln ∣ sec (θ) - 1 ∣

解答补常数

= - ln ∣ sec (θ) ∣ + ln ∣ sec (θ) - 1 ∣ + C

\int (a + x - x^{2}) d x

\int 7 x^{\frac{2}{5}} + 8 x^{- \frac{4}{5}} d x

\frac{\partial}{\partial x} (\frac{( 2 x - y )}{5 x + 3 y})

x^{2} y^{^{''}} + x y^{^{'}} - y = ln (x)

\int \frac{x}{1 - 2 x ^{2}} d x