What is the integral of-1/k cos(kx) ?

The integral of-1/k cos(kx) is -1/(k^2)sin(kx)+C

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integral of-1/k cos(kx)

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int \frac{cos ( u )}{k} d u

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

u = k x

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

Simplify

- \frac{1}{k} \cdot \frac{1}{k} sin (k x) : - \frac{1}{k ^{2}} sin (k x)

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

Add a constant to the solution

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

y = x^{2} - 35, y = 13 - 2 x

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x \to 0 - lim (\frac{x}{∣ x ∣})

\int_{1}^{2} \frac{5 x ^{2} - 1}{x} d x

f (x) = x^{2} + 1.8 x - 0.1404

What is the integral of-1/k cos(kx) ?
The integral of-1/k cos(kx) is -1/(k^2)sin(kx)+C