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受欢迎的微积分 >

integral of 3tan^3(2x)sec^5(2x)

解答

\int 3 tan^{3} (2 x) sec^{5} (2 x) d x

解答

3 (\frac{sec ^{7} ( 2 x )}{14} - \frac{sec ^{5} ( 2 x )}{10}) + C

求解步骤

\int 3 tan^{3} (2 x) sec^{5} (2 x) d x

提出常数:

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

= 3 \cdot \int tan^{3} (2 x) sec^{5} (2 x) d x

使用三角恒等式改写

= 3 \cdot \int (- 1 + sec^{2} (2 x)) tan (2 x) sec^{5} (2 x) d x

使用换元积分法

= 3 \cdot \int \frac{u ^{4} ( u ^{2} - 1 )}{2} d u

乘开

\frac{u ^{4} ( u ^{2} - 1 )}{2} : \frac{u ^{6}}{2} - \frac{u ^{4}}{2}

= 3 \cdot \int \frac{u ^{6}}{2} - \frac{u ^{4}}{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

= 3 (\int \frac{u ^{6}}{2} d u - \int \frac{u ^{4}}{2} d u)

\int \frac{u ^{6}}{2} d u = \frac{u ^{7}}{14}

\int \frac{u ^{4}}{2} d u = \frac{u ^{5}}{10}

= 3 (\frac{u ^{7}}{14} - \frac{u ^{5}}{10})

u = sec (2 x)

代回

= 3 (\frac{sec ^{7} ( 2 x )}{14} - \frac{sec ^{5} ( 2 x )}{10})

解答补常数

= 3 (\frac{sec ^{7} ( 2 x )}{14} - \frac{sec ^{5} ( 2 x )}{10}) + C

作图

流行的例子

x^{3} \cdot y^{^{'}} = x

泰勒 [(z-1)^3(z+1)^2]

t a y l or [(z - 1)^{3} (z + 1)^{2}]

(\partial)/(\partial y)(sqrt(y+(cos(x))^2))

\frac{\partial}{\partial y} (y + (cos (x))^{2})

(((1+e^x)/2)^9)^'

((\frac{1 + e ^{x}}{2})^{9})^{^{'}}

derivative of 5\sqrt[5]{x^4}

\frac{d}{d x} (5 5 x^{4})