What is the integral of cos(kx) ?

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

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integral of cos(kx)

\int cos (k x) d x

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

Solution steps

\int cos (k x) d x

Apply u-substitution

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

= \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 \int cos (u) d u

Use the common integral:

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

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

Substitute back

u = k x

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

Add a constant to the solution

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

2 x y^{^{'}} + y^{3} e^{- 2 x} = 2 x y

x^{2} + x y - y^{2} = 31, (5, 2)

\frac{\partial}{\partial x} (\frac{5 ( 1 - x )}{e ^{x}})

\frac{d}{d x} (1 + 3 cos (2 x))

(ln (x^{2} - 1))^{^{'}}