What is the integral of (-1)/2 cos(2x) ?

The integral of (-1)/2 cos(2x) is -1/4 sin(2x)+C

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integral of (-1)/2 cos(2x)

\int \frac{- 1}{2} cos (2 x) d x

- \frac{1}{4} sin (2 x) + C

Solution steps

\int \frac{- 1}{2} cos (2 x) d x

Take the constant out:

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

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

Simplify

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

Apply u-substitution

: \int cos (u) \frac{1}{2} d u

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

u = 2 x

= (- \frac{1}{2}) \frac{1}{2} sin (2 x)

Simplify

(- \frac{1}{2}) \frac{1}{2} sin (2 x) : - \frac{1}{4} sin (2 x)

= - \frac{1}{4} sin (2 x)

Add a constant to the solution

= - \frac{1}{4} sin (2 x) + C

\frac{d}{d x} (\frac{x}{z ( x )})

\frac{\partial}{\partial x} (14 x y e^{- x^{2} - y^{2}})

f (x) = x + 3, at x = 6

y^{^{''}} + 16 y = 6 tan (4 x)

\frac{d y}{d x}, x e^{y} - 10 x + 3 y = 0

What is the integral of (-1)/2 cos(2x) ?
The integral of (-1)/2 cos(2x) is -1/4 sin(2x)+C