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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

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

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

Substitute back

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