What is the integral of sin(ln(13x)) ?

The integral of sin(ln(13x)) is 1/2 x(sin(ln(13x))-cos(ln(13x)))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of sin(ln(13x))

\int sin (ln (13 x)) d x

\frac{1}{2} x (sin (ln (13 x)) - cos (ln (13 x))) + C

Solution steps

\int sin (ln (13 x)) d x

Apply u-substitution

: \int sin (ln (u)) \frac{1}{13} d u

= \int sin (ln (u)) \frac{1}{13} d u

Take the constant out:

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

= \frac{1}{13} \cdot \int sin (ln (u)) d u

Apply Integration By Parts

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

= \frac{1}{13} (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}{13} (u sin (ln (u)) - (\frac{1}{2} u sin (ln (u)) + \frac{1}{2} u cos (ln (u))))

Substitute back

u = 13 x

= \frac{1}{13} (13 x sin (ln (13 x)) - (\frac{1}{2} \cdot 13 x sin (ln (13 x)) + \frac{1}{2} \cdot 13 x cos (ln (13 x))))

Simplify

\frac{1}{13} (13 x sin (ln (13 x)) - (\frac{1}{2} \cdot 13 x sin (ln (13 x)) + \frac{1}{2} \cdot 13 x cos (ln (13 x)))) : \frac{1}{2} x (sin (ln (13 x)) - cos (ln (13 x)))

= \frac{1}{2} x (sin (ln (13 x)) - cos (ln (13 x)))

Add a constant to the solution

= \frac{1}{2} x (sin (ln (13 x)) - cos (ln (13 x))) + C

f (x) = 2 x^{2} + 5 x, at x = - 3

\frac{d ^{2} h}{d t ^{2}} = - kh

x \to 2 lim (2 x^{2} - 3 x + 1)

\int \frac{x ^{2}}{4} - 1 d x

y^{^{''}} + 36 y = - 36 sec (6 t)

What is the integral of sin(ln(13x)) ?
The integral of sin(ln(13x)) is 1/2 x(sin(ln(13x))-cos(ln(13x)))+C