integral of \sqrt[5]{tan(x)}sec^4(x)

解答

\int 5 tan (x) sec^{4} (x) d x

\frac{5}{6} tan^{\frac{6}{5}} (x) + \frac{5}{16} tan^{\frac{16}{5}} (x) + C

求解步骤

\int 5 tan (x) sec^{4} (x) d x

使用三角恒等式改写

= \int (1 + tan^{2} (x)) sec^{2} (x) 5 tan (x) d x

使用换元积分法

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

乘开

5 u (1 + u^{2}) : 5 u + u^{\frac{11}{5}}

= \int 5 u + u^{\frac{11}{5}} 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 5 u d u + \int u^{\frac{11}{5}} d u

\int 5 u d u = \frac{5}{6} u^{\frac{6}{5}}

\int u^{\frac{11}{5}} d u = \frac{5}{16} u^{\frac{16}{5}}

= \frac{5}{6} u^{\frac{6}{5}} + \frac{5}{16} u^{\frac{16}{5}}

u = tan (x)

代回

= \frac{5}{6} tan^{\frac{6}{5}} (x) + \frac{5}{16} tan^{\frac{16}{5}} (x)

解答补常数

= \frac{5}{6} tan^{\frac{6}{5}} (x) + \frac{5}{16} tan^{\frac{16}{5}} (x) + C

\int sin (x) e^{x} d x

d er i v a t i v e e^{5 t}

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

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

\frac{\partial}{\partial x} (2 x^{2} y - y^{3} + x)