What is the integral of (cos(ln(9t)))/t ?

The integral of (cos(ln(9t)))/t is sin(ln(9t))+C

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integral of (cos(ln(9t)))/t

\int \frac{cos ( ln ( 9 t ) )}{t} d t

sin (ln (9 t)) + C

Solution steps

\int \frac{cos ( ln ( 9 t ) )}{t} d t

Apply u-substitution

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

= \int \frac{cos ( ln ( u ) )}{u} d u

Apply u-substitution

: \int cos (v) d v

= \int cos (v) d v

Use the common integral:

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

= sin (v)

Substitute back

= sin (ln (9 t))

Add a constant to the solution

= sin (ln (9 t)) + C

\frac{d}{d x} (e^{2 c o s (3 x)})

y^{^{''}} + 4 y^{^{'}} + 4 y = 0, y (0) = 2, y^{^{'}} (0) = - 9

y^{2} - 6 x = 1, x - y = 1

\int - e^{- 2 x} x d x

\int (e) d x

What is the integral of (cos(ln(9t)))/t ?
The integral of (cos(ln(9t)))/t is sin(ln(9t))+C