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

解答

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

\frac{2}{3} tan^{\frac{3}{2}} (x) + \frac{2}{7} tan^{\frac{7}{2}} (x) + C

求解步骤

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

使用三角恒等式改写

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

使用换元积分法

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

乘开

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

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

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

\int u^{\frac{5}{2}} d u = \frac{2}{7} u^{\frac{7}{2}}

= \frac{2}{3} u^{\frac{3}{2}} + \frac{2}{7} u^{\frac{7}{2}}

u = tan (x)

代回

= \frac{2}{3} tan^{\frac{3}{2}} (x) + \frac{2}{7} tan^{\frac{7}{2}} (x)

解答补常数

= \frac{2}{3} tan^{\frac{3}{2}} (x) + \frac{2}{7} tan^{\frac{7}{2}} (x) + C

\int_{5}^{1} - 4 x + 1 d x

n = 0 \sum \infty - 2 - 8 - 32 - 128

x \to 3 lim (\frac{x ^{2} + 5}{x + 5})

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

t an g e n t y = x^{2} - 1, (- 3, 8)