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

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

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integral of-cos(4x)

\int - cos (4 x) d x

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

Solution steps

\int - cos (4 x) d x

Take the constant out:

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

= - \int cos (4 x) d x

Apply u-substitution

: \int cos (u) \frac{1}{4} d u

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

u = 4 x

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

Add a constant to the solution

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

\frac{d}{d x} (7^{x + a r c c o s (x)})

y^{^{''}} + 4 y^{^{'}} + 20 y = 0

\frac{\partial}{\partial x} (0.0099999)

\int 1 + 5 x^{4} (20 x^{3}) d x

n = 1 \sum \infty n (\frac{3}{2})^{n}