What is the integral of sin(ln(3x)) ?

The integral of sin(ln(3x)) is 1/2 x(sin(ln(3x))-cos(ln(3x)))+C

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integral of sin(ln(3x))

\int sin (ln (3 x)) d x

\frac{1}{2} x (sin (ln (3 x)) - cos (ln (3 x))) + C

Solution steps

\int sin (ln (3 x)) d x

Apply u-substitution

: \int sin (ln (u)) \frac{1}{3} d u

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

Apply Integration By Parts

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

= \frac{1}{3} (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}{3} (u sin (ln (u)) - (\frac{1}{2} u sin (ln (u)) + \frac{1}{2} u cos (ln (u))))

Substitute back

u = 3 x

= \frac{1}{3} (3 x sin (ln (3 x)) - (\frac{1}{2} \cdot 3 x sin (ln (3 x)) + \frac{1}{2} \cdot 3 x cos (ln (3 x))))

Simplify

\frac{1}{3} (3 x sin (ln (3 x)) - (\frac{1}{2} \cdot 3 x sin (ln (3 x)) + \frac{1}{2} \cdot 3 x cos (ln (3 x)))) : \frac{1}{2} x (sin (ln (3 x)) - cos (ln (3 x)))

= \frac{1}{2} x (sin (ln (3 x)) - cos (ln (3 x)))

Add a constant to the solution

= \frac{1}{2} x (sin (ln (3 x)) - cos (ln (3 x))) + C

\frac{d}{d x} (\frac{4}{x ^{2} + 2})

\int_{0}^{l n (2)} e^{2 x} d x

\int \frac{9}{2} (t + 1)^{\frac{1}{2}} d t

y^{^{'}} = x^{3} (1 - y), y (0) = 3

\int \frac{x - 1}{x + 1} d x

What is the integral of sin(ln(3x)) ?
The integral of sin(ln(3x)) is 1/2 x(sin(ln(3x))-cos(ln(3x)))+C