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 1/(1+sin(x)cos(x))

解答

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

解答

\frac{2}{3} arctan (\frac{1}{3} (2 tan (x) + 1)) + C

求解步骤

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

使用三角换元法

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

对

1 + u^{2} + u

配方:

(u + \frac{1}{2})^{2} + \frac{3}{4}

= \int \frac{1}{( u + \frac{1}{2} ) ^{2} + \frac{3}{4}} d u

使用换元积分法

= \int \frac{4}{4 v ^{2} + 3} d v

提出常数:

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

= 4 \cdot \int \frac{1}{4 v ^{2} + 3} d v

使用换元积分法

= 4 \cdot \int \frac{1}{2 3 ( w ^{2} + 1 )} d w

提出常数:

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

= 4 \cdot \frac{1}{2 3} \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 = arctan (w)

= 4 \cdot \frac{1}{2 3} arctan (w)

代回

= 4 \cdot \frac{1}{2 3} arctan (\frac{2}{3} (tan (x) + \frac{1}{2}))

化简

4 \cdot \frac{1}{2 3} arctan (\frac{2}{3} (tan (x) + \frac{1}{2})) : \frac{2}{3} arctan (\frac{1}{3} (2 tan (x) + 1))

= \frac{2}{3} arctan (\frac{1}{3} (2 tan (x) + 1))

解答补常数

= \frac{2}{3} arctan (\frac{1}{3} (2 tan (x) + 1)) + C

作图

流行的例子

麦克劳林 (1+x)^{-6/5}

ma c l a u r in (1 + x)^{- \frac{6}{5}}

tangent y=(1-3x)^{10},(0,1)

t an g e n t y = (1 - 3 x)^{10}, (0, 1)

integral of e^{5-x}

\int e^{5 - x} d x

integral from 0 to 1 of (2x-1)^6

\int_{0}^{1} (2 x - 1)^{6} d x

integral of 1/((5x-2)^{3/2)}

\int \frac{1}{( 5 x - 2 ) ^{\frac{3}{2}}} d x