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

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

Use the common integral:

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

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

Substitute back

u = 2 x

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

Simplify

- \frac{1}{2} \cdot \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{( 2 s + 1 )}{( s ^{2} - 2 s + 2 )}

\frac{\partial}{\partial x} ((y^{2}) d x)

(2 x + 3)^{2}

4 x^{2} (\frac{d y}{d x}) = (\frac{d y}{d x}) + 3 x e^{- y}

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

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