zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的微积分 >

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

解答

\int tan (x) (sec (x) - tan (x)) d x

解答

4 (\frac{1}{2} arctan (tan (\frac{x}{2})) + \frac{1}{2 ( tan ( \frac{x}{2} ) + 1 )}) + C

求解步骤

\int tan (x) (sec (x) - tan (x)) d x

使用三角恒等式改写

= \int \frac{sin ( x ) ( 1 - sin ( x ) )}{cos ^{2} ( x )} d x

使用三角恒等式改写

= \int \frac{1}{1 + csc ( x )} d x

使用换元积分法

= \int \frac{4 u}{( 2 u + 1 + u ^{2} ) ( 1 + u ^{2} )} d u

提出常数:

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

= 4 \cdot \int \frac{u}{( 2 u + 1 + u ^{2} ) ( 1 + u ^{2} )} d u

使用分布积分法

= 4 (u (\frac{1}{2} ln ∣ u + 1 ∣ - \frac{1}{2 ( u + 1 )} - \frac{1}{4} ln u^{2} + 1) - \int \frac{1}{2} ln ∣ u + 1 ∣ - \frac{1}{2 ( u + 1 )} - \frac{1}{4} ln u^{2} + 1 d u)

\int \frac{1}{2} ln ∣ u + 1 ∣ - \frac{1}{2 ( u + 1 )} - \frac{1}{4} ln u^{2} + 1 d u = \frac{1}{2} u ln ∣ u + 1 ∣ - \frac{u}{2 ( u + 1 )} - \frac{1}{4} u ln u^{2} + 1 - \frac{1}{2 ( u + 1 )} - \frac{1}{2} arctan (u)

= 4 (u (\frac{1}{2} ln ∣ u + 1 ∣ - \frac{1}{2 ( u + 1 )} - \frac{1}{4} ln u^{2} + 1) - (\frac{1}{2} u ln ∣ u + 1 ∣ - \frac{u}{2 ( u + 1 )} - \frac{1}{4} u ln u^{2} + 1 - \frac{1}{2 ( u + 1 )} - \frac{1}{2} arctan (u)))

u = tan (\frac{x}{2})

代回

= 4 (tan (\frac{x}{2}) (\frac{1}{2} ln tan (\frac{x}{2}) + 1 - \frac{1}{2 ( tan ( \frac{x}{2} ) + 1 )} - \frac{1}{4} ln tan^{2} (\frac{x}{2}) + 1) - (\frac{1}{2} tan (\frac{x}{2}) ln tan (\frac{x}{2}) + 1 - \frac{tan ( \frac{x}{2} )}{2 ( tan ( \frac{x}{2} ) + 1 )} - \frac{1}{4} tan (\frac{x}{2}) ln tan^{2} (\frac{x}{2}) + 1 - \frac{1}{2 ( tan ( \frac{x}{2} ) + 1 )} - \frac{1}{2} arctan (tan (\frac{x}{2}))))

化简

4 (tan (\frac{x}{2}) (\frac{1}{2} ln tan (\frac{x}{2}) + 1 - \frac{1}{2 ( tan ( \frac{x}{2} ) + 1 )} - \frac{1}{4} ln tan^{2} (\frac{x}{2}) + 1) - (\frac{1}{2} tan (\frac{x}{2}) ln tan (\frac{x}{2}) + 1 - \frac{tan ( \frac{x}{2} )}{2 ( tan ( \frac{x}{2} ) + 1 )} - \frac{1}{4} tan (\frac{x}{2}) ln tan^{2} (\frac{x}{2}) + 1 - \frac{1}{2 ( tan ( \frac{x}{2} ) + 1 )} - \frac{1}{2} arctan (tan (\frac{x}{2})))) : 4 (\frac{1}{2} arctan (tan (\frac{x}{2})) + \frac{1}{2 ( tan ( \frac{x}{2} ) + 1 )})

= 4 (\frac{1}{2} arctan (tan (\frac{x}{2})) + \frac{1}{2 ( tan ( \frac{x}{2} ) + 1 )})

解答补常数

= 4 (\frac{1}{2} arctan (tan (\frac{x}{2})) + \frac{1}{2 ( tan ( \frac{x}{2} ) + 1 )}) + C

作图

流行的例子

integral of zln(z)

\int z ln (z) d z

integral of x^{-1}*1/(2+x^2)

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

integral of (sec(x))^2(tan(x))^2

\int (sec (x))^{2} (tan (x))^{2} d x

integral of x^{23}

\int x^{23} d x

integral of (10x^4-18x^2+2x+3)

\int (10 x^{4} - 18 x^{2} + 2 x + 3) d x