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受欢迎的微积分 >

integral of (sec(x))/(1-tan(x))

解答

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

解答

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

求解步骤

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

用 sin, cos 表示

= \int \frac{1}{cos ( x ) - sin ( x )} d x

使用换元积分法

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

提出常数:

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

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

对

1 - u^{2} - 2 u

配方:

- (u + 1)^{2} + 2

= 2 \cdot \int \frac{1}{- ( u + 1 ) ^{2} + 2} d u

使用换元积分法

= 2 \cdot \int \frac{1}{- v ^{2} + 2} d v

使用换元积分法

= 2 \cdot \int \frac{1}{2 ( - w ^{2} + 1 )} d w

提出常数:

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

= 2 \cdot \frac{1}{2} \cdot \int \frac{1}{- w ^{2} + 1} d w

使用积分常见定则:

\int \frac{1}{- w ^{2} + 1} d w = \frac{ln ∣ w + 1 ∣}{2} - \frac{ln ∣ w - 1 ∣}{2}

= 2 \cdot \frac{1}{2} (\frac{ln ∣ w + 1 ∣}{2} - \frac{ln ∣ w - 1 ∣}{2})

代回

= 2 \cdot \frac{1}{2} \frac{ln \frac{t a n ( \frac{x}{2} ) + 1}{2} + 1}{2} - \frac{ln \frac{t a n ( \frac{x}{2} ) + 1}{2} - 1}{2}

化简

2 \cdot \frac{1}{2} \frac{ln \frac{t a n ( \frac{x}{2} ) + 1}{2} + 1}{2} - \frac{ln \frac{t a n ( \frac{x}{2} ) + 1}{2} - 1}{2} : 2 (\frac{1}{2} ln \frac{tan ( \frac{x}{2} ) + 1}{2} + 1 - \frac{1}{2} ln \frac{tan ( \frac{x}{2} ) + 1}{2} - 1)

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

解答补常数

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

作图

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