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受欢迎的微积分 >

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

解答

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

解答

\frac{1}{9} tan^{\frac{3}{2}} (6 x) + C

求解步骤

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

使用三角恒等式改写

= \int \frac{sin ( 6 x )}{cos ( 6 x )} sec^{2} (6 x) d x

使用基本三角恒等式:

\frac{sin ( x )}{cos ( x )} = tan (x)

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

使用换元积分法

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

提出常数:

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

= \frac{1}{3} \cdot \int u^{2} d u

使用幂法则

= \frac{1}{3} \cdot \frac{u ^{3}}{3}

u = tan (6 x)

代回

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

化简

\frac{1}{3} \cdot \frac{( tan ( 6 x ) ) ^{3}}{3} : \frac{1}{9} tan^{\frac{3}{2}} (6 x)

= \frac{1}{9} tan^{\frac{3}{2}} (6 x)

解答补常数

= \frac{1}{9} tan^{\frac{3}{2}} (6 x) + C

作图

流行的例子

integral of 8sin(sqrt(x))

\int 8 sin (x) d x

d/(dy)(1-3/y+x)

\frac{d}{d y} (1 - \frac{3}{y} + x)

(dy)/(dx)=-y^2sin(x)

\frac{d y}{d x} = - y^{2} sin (x)

integral of 1/(x(x^2-49)^{3/2)}

\int \frac{1}{x ( x ^{2} - 49 ) ^{\frac{3}{2}}} d x

(y^2+yx)dx-x^2dy=0

(y^{2} + y x) d x - x^{2} d y = 0