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受欢迎的微积分 >

integral of (-16cos(2t))

解答

\int (- 16 cos (2 t)) d t

解答

- 8 sin (2 t) + C

求解步骤

\int - 16 cos (2 t) d t

提出常数:

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

= - 16 \cdot \int cos (2 t) d t

使用换元积分法

= - 16 \cdot \int cos (u) \frac{1}{2} d u

提出常数:

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

= - 16 \cdot \frac{1}{2} \cdot \int cos (u) d u

使用积分常见定则:

\int cos (u) d u = sin (u)

= - 16 \cdot \frac{1}{2} sin (u)

u = 2 t

代回

= - 16 \cdot \frac{1}{2} sin (2 t)

化简

- 16 \cdot \frac{1}{2} sin (2 t) : - 8 sin (2 t)

= - 8 sin (2 t)

解答补常数

= - 8 sin (2 t) + C

作图

流行的例子

integral of ((sin(x))/(cos(x)))

\int (\frac{sin ( x )}{cos ( x )}) d x

derivative of (sqrt(y-x^2+2)/(x-y))

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

y^'=(y^2+2)/(xy-4y)

y^{^{'}} = \frac{y ^{2} + 2}{x y - 4 y}

(\partial)/(\partial x)(xye^{4y})

\frac{\partial}{\partial x} (x y e^{4 y})

tangent y=x^2+6x-5,\at x=2

t an g e n t y = x^{2} + 6 x - 5, at x = 2