What is the integral of sin(ln(9x)) ?

The integral of sin(ln(9x)) is 1/2 x(sin(ln(9x))-cos(ln(9x)))+C

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integral of sin(ln(9x))

\int sin (ln (9 x)) d x

\frac{1}{2} x (sin (ln (9 x)) - cos (ln (9 x))) + C

Solution steps

\int sin (ln (9 x)) d x

Apply u-substitution

: \int sin (ln (u)) \frac{1}{9} d u

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

Apply Integration By Parts

: u sin (ln (u)) - \int cos (ln (u)) d u

= \frac{1}{9} (u sin (ln (u)) - \int cos (ln (u)) d u)

\int cos (ln (u)) d u = \frac{1}{2} u sin (ln (u)) + \frac{1}{2} u cos (ln (u))

= \frac{1}{9} (u sin (ln (u)) - (\frac{1}{2} u sin (ln (u)) + \frac{1}{2} u cos (ln (u))))

Substitute back

u = 9 x

= \frac{1}{9} (9 x sin (ln (9 x)) - (\frac{1}{2} \cdot 9 x sin (ln (9 x)) + \frac{1}{2} \cdot 9 x cos (ln (9 x))))

Simplify

\frac{1}{9} (9 x sin (ln (9 x)) - (\frac{1}{2} \cdot 9 x sin (ln (9 x)) + \frac{1}{2} \cdot 9 x cos (ln (9 x)))) : \frac{1}{2} x (sin (ln (9 x)) - cos (ln (9 x)))

= \frac{1}{2} x (sin (ln (9 x)) - cos (ln (9 x)))

Add a constant to the solution

= \frac{1}{2} x (sin (ln (9 x)) - cos (ln (9 x))) + C

\frac{d y}{d x} + 2 y = 5

ln (2 - x^{2} + 2 x^{4})

y^{^{'}} = 2 cos (t) + \frac{y}{2}

x e^{2 x}

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

What is the integral of sin(ln(9x)) ?
The integral of sin(ln(9x)) is 1/2 x(sin(ln(9x))-cos(ln(9x)))+C