What is the integral of sin(kt) ?

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

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integral of sin(kt)

\int sin (k t) d t

- \frac{1}{k} cos (k t) + C

Solution steps

\int sin (k t) d t

Apply u-substitution

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

= \int \frac{sin ( 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 sin (u) d u

Use the common integral:

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

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

Substitute back

u = k t

= \frac{1}{k} (- cos (k t))

Simplify

= - \frac{1}{k} cos (k t)

Add a constant to the solution

= - \frac{1}{k} cos (k t) + C

- t^{2} + 9 t - 20

\int \frac{6 x ^{2}}{( 2 x - 1 ) ( 2 x + 3 )} d x

(x^{2} - 3 x y) d x + x d y = 0

\frac{d ^{2} y}{d t ^{2}} + 4 \frac{d y}{d t} - 5 y = 4

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