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受欢迎的微积分 >

integral of ((sec(x))/(1+tan(x)))^2

解答

\int (\frac{sec ( x )}{1 + tan ( x )})^{2} d x

解答

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

求解步骤

\int (\frac{sec ( x )}{1 + tan ( x )})^{2} d x

使用三角恒等式改写

= \int \frac{sec ^{2} ( x ) cos ^{2} ( x )}{( cos ( x ) + sin ( x ) ) ^{2}} d x

化简

\frac{sec ^{2} ( x ) cos ^{2} ( x )}{( cos ( x ) + sin ( x ) ) ^{2}} : \frac{1}{1 + sin ( 2 x )}

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

使用换元积分法

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

提出常数:

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

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

使用换元积分法

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

提出常数:

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

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

对

1 + v^{2} + 2 v

配方:

(v + 1)^{2}

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

使用换元积分法

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

使用幂法则

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

代回

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

化简

\frac{1}{2} \cdot 2 (- \frac{1}{tan ( \frac{2 x}{2} ) + 1}) : - \frac{1}{tan ( x ) + 1}

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

解答补常数

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

作图

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