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受欢迎的微积分 >

integral of tan^2(6x)sec^4(6x)

解答

\int tan^{2} (6 x) sec^{4} (6 x) d x

解答

\frac{1}{6} (\frac{tan ^{3} ( 6 x )}{3} + \frac{tan ^{5} ( 6 x )}{5}) + C

求解步骤

\int tan^{2} (6 x) sec^{4} (6 x) d x

使用换元积分法

= \int tan^{2} (u) sec^{4} (u) \frac{1}{6} d u

提出常数:

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

= \frac{1}{6} \cdot \int tan^{2} (u) sec^{4} (u) d u

使用三角恒等式改写

= \frac{1}{6} \cdot \int (1 + tan^{2} (u)) sec^{2} (u) tan^{2} (u) d u

使用换元积分法

= \frac{1}{6} \cdot \int v^{2} (1 + v^{2}) d v

乘开

v^{2} (1 + v^{2}) : v^{2} + v^{4}

= \frac{1}{6} \cdot \int v^{2} + v^{4} d v

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

= \frac{1}{6} (\int v^{2} d v + \int v^{4} d v)

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

\int v^{4} d v = \frac{v ^{5}}{5}

= \frac{1}{6} (\frac{v ^{3}}{3} + \frac{v ^{5}}{5})

代回

= \frac{1}{6} (\frac{tan ^{3} ( 6 x )}{3} + \frac{tan ^{5} ( 6 x )}{5})

解答补常数

= \frac{1}{6} (\frac{tan ^{3} ( 6 x )}{3} + \frac{tan ^{5} ( 6 x )}{5}) + C

作图

流行的例子

面积 x^3,-x,8

a re a x^{3}, - x, 8

integral from 0 to 6 of 9sqrt(4x+1)

\int_{0}^{6} 9 4 x + 1 d x

36y^{''}-y=0,y(-5)=1,y^'(-5)=-1

36 y^{^{''}} - y = 0, y (- 5) = 1, y^{^{'}} (- 5) = - 1

y^{^{''}} - y^{^{'}} = 0

limit as x approaches 2+of tan((pix)/4)

x \to 2 + lim (tan (\frac{π x}{4}))