integral of 6tan^4(x)

解答

\int 6 tan^{4} (x) d x

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

求解步骤

\int 6 tan^{4} (x) d x

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 6 \cdot \int tan^{4} (x) d x

使用降次积分法

= 6 (\frac{tan ^{3} ( x )}{3} - \int tan^{2} (x) d x)

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

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

化简

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

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

解答补常数

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

\frac{\partial}{\partial x} (2 x + y + 1)

\int \frac{3 x ^{2} - 7}{x ^{3} - 7 x + 6} d x

\int_{0}^{1} (t 4 + 9 t^{2}) d t

s l o p e in t erce pt x + 2 y = x + 2 y = 8

y^{^{'}} = 5 x (x - 3)