What is the integral of sin(ln(2x)) ?

The integral of sin(ln(2x)) is 1/2 (xsin(ln(2x))-xcos(ln(2x)))+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(2x))

\int sin (ln (2 x)) d x

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

Solution steps

\int sin (ln (2 x)) d x

Apply u-substitution

: \int sin (ln (u)) \frac{1}{2} d u

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

Take the constant out:

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

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

Apply Integration By Parts

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

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

Substitute back

u = 2 x

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (2 sin (2 x) - 2 cos (2 x) - \frac{1}{5} e^{x})

\frac{d}{d x} ((x - 5)^{2})

n = 0 \sum \infty (- 1)^{n} \cdot n

y = - 80 + 208

\frac{\partial}{\partial y} (tan (x) y^{2.5})

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