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

The integral of (sin(9ln(t)))/t is -1/9 cos(9ln(t))+C

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

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

- \frac{1}{9} cos (9 ln (t)) + C

Solution steps

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

Apply u-substitution

: \int \frac{sin ( u )}{9} d u

= \int \frac{sin ( u )}{9} d u

Take the constant out:

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

= \frac{1}{9} \cdot \int sin (u) d u

Use the common integral:

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

= \frac{1}{9} (- cos (u))

Substitute back

u = 9 ln (t)

= \frac{1}{9} (- cos (9 ln (t)))

Simplify

= - \frac{1}{9} cos (9 ln (t))

Add a constant to the solution

= - \frac{1}{9} cos (9 ln (t)) + C

n = 1 \sum \infty arctan (6 n)

\int x^{2} x d x

\frac{d}{d t} (e^{(7 t s i n (2 t))})

e^{, at}

y^{^{'}} = 2 y - t

What is the integral of (sin(9ln(t)))/t ?
The integral of (sin(9ln(t)))/t is -1/9 cos(9ln(t))+C