What is the integral of sin(ln(12x)) ?

The integral of sin(ln(12x)) is 1/2 x(sin(ln(12x))-cos(ln(12x)))+C

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integral of sin(ln(12x))

\int sin (ln (12 x)) d x

\frac{1}{2} x (sin (ln (12 x)) - cos (ln (12 x))) + C

Solution steps

\int sin (ln (12 x)) d x

Apply u-substitution

: \int sin (ln (u)) \frac{1}{12} d u

= \int sin (ln (u)) \frac{1}{12} d u

Take the constant out:

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

= \frac{1}{12} \cdot \int sin (ln (u)) d u

Apply Integration By Parts

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

= \frac{1}{12} (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}{12} (u sin (ln (u)) - (\frac{1}{2} u sin (ln (u)) + \frac{1}{2} u cos (ln (u))))

Substitute back

u = 12 x

= \frac{1}{12} (12 x sin (ln (12 x)) - (\frac{1}{2} \cdot 12 x sin (ln (12 x)) + \frac{1}{2} \cdot 12 x cos (ln (12 x))))

Simplify

\frac{1}{12} (12 x sin (ln (12 x)) - (\frac{1}{2} \cdot 12 x sin (ln (12 x)) + \frac{1}{2} \cdot 12 x cos (ln (12 x)))) : \frac{1}{2} x (sin (ln (12 x)) - cos (ln (12 x)))

= \frac{1}{2} x (sin (ln (12 x)) - cos (ln (12 x)))

Add a constant to the solution

= \frac{1}{2} x (sin (ln (12 x)) - cos (ln (12 x))) + C

6 - x^{2}, 26 - 9 x

\frac{d}{d t} (2 sin (3 t))

\int 3 x^{2} - x + 3 d x

\frac{d}{d x} (\frac{8 x ^{2} + 2 x + 2}{x})

What is the integral of sin(ln(12x)) ?
The integral of sin(ln(12x)) is 1/2 x(sin(ln(12x))-cos(ln(12x)))+C