integral of (sec^4(x))/(sqrt(tan(x)))

解答

\int \frac{sec ^{4} ( x )}{tan ( x )} d x

2 tan (x) + \frac{2}{5} tan^{\frac{5}{2}} (x) + C

求解步骤

\int \frac{sec ^{4} ( x )}{tan ( x )} d x

使用换元积分法

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

乘开

\frac{1 + u ^{2}}{u} : \frac{1}{u} + u^{\frac{3}{2}}

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

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

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

\int u^{\frac{3}{2}} d u = \frac{2}{5} u^{\frac{5}{2}}

= 2 u + \frac{2}{5} u^{\frac{5}{2}}

u = tan (x)

代回

= 2 tan (x) + \frac{2}{5} tan^{\frac{5}{2}} (x)

解答补常数

= 2 tan (x) + \frac{2}{5} tan^{\frac{5}{2}} (x) + C

\frac{\partial}{\partial y} (7 x sin (9 x^{2} y))

x \to - \infty lim (x e^{- x + 2})

\frac{\partial}{\partial x} (ln (x + 3 y + 4 z))

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

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