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受欢迎的微积分 >

integral of (2sec(x))/(2tan(x)+3sec(x))

解答

\int \frac{2 sec ( x )}{2 tan ( x ) + 3 sec ( x )} d x

解答

\frac{4}{5} arctan (\frac{1}{5} (3 tan (\frac{x}{2}) + 2)) + C

求解步骤

\int \frac{2 sec ( x )}{2 tan ( x ) + 3 sec ( x )} d x

提出常数:

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

= 2 \cdot \int \frac{sec ( x )}{2 tan ( x ) + 3 sec ( x )} d x

使用三角恒等式改写

= 2 \cdot \int \frac{1}{2 sin ( x ) + 3} d x

使用换元积分法

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

提出常数:

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

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

乘开

3 (1 + u^{2}) : 3 + 3 u^{2}

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

对

4 u + 3 + 3 u^{2}

配方:

3 (u + \frac{2}{3})^{2} + \frac{5}{3}

= 2 \cdot 2 \cdot \int \frac{1}{3 ( u + \frac{2}{3} ) ^{2} + \frac{5}{3}} d u

使用换元积分法

= 2 \cdot 2 \cdot \int \frac{3}{9 v ^{2} + 5} d v

提出常数:

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

= 2 \cdot 2 \cdot 3 \cdot \int \frac{1}{9 v ^{2} + 5} d v

使用换元积分法

= 2 \cdot 2 \cdot 3 \cdot \int \frac{1}{3 5 ( 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 2 \cdot 3 \cdot \frac{1}{3 5} \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)

= 2 \cdot 2 \cdot 3 \cdot \frac{1}{3 5} arctan (w)

代回

= 2 \cdot 2 \cdot 3 \cdot \frac{1}{3 5} arctan (\frac{3}{5} (tan (\frac{x}{2}) + \frac{2}{3}))

化简

2 \cdot 2 \cdot 3 \cdot \frac{1}{3 5} arctan (\frac{3}{5} (tan (\frac{x}{2}) + \frac{2}{3})) : \frac{4}{5} arctan (\frac{1}{5} (3 tan (\frac{x}{2}) + 2))

= \frac{4}{5} arctan (\frac{1}{5} (3 tan (\frac{x}{2}) + 2))

解答补常数

= \frac{4}{5} arctan (\frac{1}{5} (3 tan (\frac{x}{2}) + 2)) + C

作图

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