integral of 1/((sin(x)+tan(x)))

解答

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

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

求解步骤

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

使用换元积分法

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

乘开

\frac{( u + 1 ) ( u - 1 )}{u} : u - \frac{1}{u}

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

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

\int \frac{1}{u} d u = ln ∣ u ∣

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

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

代回

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

解答补常数

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

\frac{\partial}{\partial x} (1 - y^{2})

\int \frac{1}{- x ^{2} - 4 x} d x

h \to 0 lim (\frac{( 1 + h ) ^{8} - 1}{h})

\int ln (2 x + 3) d x

\int \frac{5 - x ^{2}}{( x ^{2} + 9 ) ( 3 - x )} d x