What is the integral of-(cos(2x))/4 ?

The integral of-(cos(2x))/4 is -1/8 sin(2x)+C

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integral of-(cos(2x))/4

\int - \frac{cos ( 2 x )}{4} d x

- \frac{1}{8} sin (2 x) + C

Solution steps

\int - \frac{cos ( 2 x )}{4} d x

Take the constant out:

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

= - \frac{1}{4} \cdot \int cos (2 x) d x

Apply u-substitution

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

Use the common integral:

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

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

Substitute back

u = 2 x

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

Simplify

- \frac{1}{4} \cdot \frac{1}{2} sin (2 x) : - \frac{1}{8} sin (2 x)

= - \frac{1}{8} sin (2 x)

Add a constant to the solution

= - \frac{1}{8} sin (2 x) + C

\frac{d y}{d x} = \frac{2}{3} y

x \to 4 lim (x + 1)

\frac{\partial}{\partial x} (ln (4 x^{2} + y^{2} + 8))

\frac{d}{d x} ((4 x^{\frac{3}{2}} - 2 x^{\frac{1}{2}})^{3})

w (z) = 7 - z

What is the integral of-(cos(2x))/4 ?
The integral of-(cos(2x))/4 is -1/8 sin(2x)+C