integral of tan^4(2x)

解答

\int tan^{4} (2 x) d x

\frac{1}{6} tan^{3} (2 x) + x - \frac{1}{2} tan (2 x) + C

求解步骤

\int tan^{4} (2 x) d x

使用换元积分法

= \int tan^{4} (u) \frac{1}{2} 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 tan^{4} (u) d u

使用降次积分法

= \frac{1}{2} (\frac{tan ^{3} ( u )}{3} - \int tan^{2} (u) d u)

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

= \frac{1}{2} (\frac{tan ^{3} ( u )}{3} - (- u + tan (u)))

u = 2 x

代回

= \frac{1}{2} (\frac{tan ^{3} ( 2 x )}{3} - (- 2 x + tan (2 x)))

化简

\frac{1}{2} (\frac{tan ^{3} ( 2 x )}{3} - (- 2 x + tan (2 x))) : \frac{1}{6} tan^{3} (2 x) + x - \frac{1}{2} tan (2 x)

= \frac{1}{6} tan^{3} (2 x) + x - \frac{1}{2} tan (2 x)

解答补常数

= \frac{1}{6} tan^{3} (2 x) + x - \frac{1}{2} tan (2 x) + C

t^{5} y^{4} - t^{3} y^{^{'}} + 6 y = 0

y^{^{'}} = \frac{3 ( e ^{x} + 1 )}{2 y}

s im pl i f y \frac{d}{dd} (\frac{a x + b}{c x + d})

\frac{d}{d x} (- \frac{1}{2} x^{4} cos (x))

\frac{d}{d x} (3 x ln (x))