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

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

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integral of-1/4 cos(2x)

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

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

Solution steps

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

Apply u-substitution

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

= - \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

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\int \frac{x}{t} d t

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

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