What is the integral of-1/4 sin(2x) ?

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

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integral of-1/4 sin(2x)

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

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

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

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

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

Substitute back

u = 2 x

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

Simplify

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

= \frac{1}{8} cos (2 x)

Add a constant to the solution

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

f (x) = \frac{6 x}{x + 6}

\frac{d}{d x} (ln (3 x^{2} + 5))

36 y^{^{''}} - 12 y^{^{'}} + y = 0

x \to 0 + lim (x^{2} - ln (x))

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

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