What is the integral of (cos(sqrt(6t)))/(sqrt(6t)) ?

The integral of (cos(sqrt(6t)))/(sqrt(6t)) is 1/3 sin(sqrt(6)sqrt(t))+C

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integral of (cos(sqrt(6t)))/(sqrt(6t))

\int \frac{cos ( 6 t )}{6 t} d t

\frac{1}{3} sin (6 t) + C

Solution steps

\int \frac{cos ( 6 t )}{6 t} d t

Simplify

\frac{cos ( 6 t )}{6 t} : \frac{cos ( 6 t )}{6 t}

= \int \frac{cos ( 6 t )}{6 t} d t

Take the constant out:

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

= \frac{1}{6} \cdot \int \frac{cos ( 6 t )}{t} d t

Apply u-substitution

: \int \frac{2 cos ( u )}{3} d u

= \frac{1}{6} \cdot \int \frac{2 cos ( u )}{3} d u

Take the constant out:

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

= \frac{1}{6} \cdot \frac{2}{3} \cdot \int cos (u) d u

Combine same powers :

\frac{x}{y} = \frac{x}{y}

= \frac{1}{6} \frac{2}{3} \cdot \int cos (u) d u

Use the common integral:

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

= \frac{1}{6} \frac{2}{3} sin (u)

Substitute back

u = 6 t

= \frac{1}{6} \frac{2}{3} sin (6 t)

Simplify

\frac{1}{6} \frac{2}{3} sin (6 t) : \frac{1}{3} sin (6 t)

= \frac{1}{3} sin (6 t)

Add a constant to the solution

= \frac{1}{3} sin (6 t) + C

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What is the integral of (cos(sqrt(6t)))/(sqrt(6t)) ?
The integral of (cos(sqrt(6t)))/(sqrt(6t)) is 1/3 sin(sqrt(6)sqrt(t))+C