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

The integral of sin^2(ln(x)) is 1/10 (5x-2xsin(2ln(x))-xcos(2ln(x)))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of sin^2(ln(x))

\int sin^{2} (ln (x)) d x

\frac{1}{10} (5 x - 2 x sin (2 ln (x)) - x cos (2 ln (x))) + C

Solution steps

\int sin^{2} (ln (x)) d x

Rewrite using trig identities

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

Take the constant out:

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

= \frac{1}{2} \cdot \int 1 - cos (2 ln (x)) d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int 1 d x = x

\int cos (2 ln (x)) d x = \frac{2}{5} x sin (2 ln (x)) + \frac{1}{5} x cos (2 ln (x))

= \frac{1}{2} (x - (\frac{2}{5} x sin (2 ln (x)) + \frac{1}{5} x cos (2 ln (x))))

Simplify

\frac{1}{2} (x - (\frac{2}{5} x sin (2 ln (x)) + \frac{1}{5} x cos (2 ln (x)))) : \frac{1}{10} (5 x - 2 x sin (2 ln (x)) - x cos (2 ln (x)))

= \frac{1}{10} (5 x - 2 x sin (2 ln (x)) - x cos (2 ln (x)))

Add a constant to the solution

= \frac{1}{10} (5 x - 2 x sin (2 ln (x)) - x cos (2 ln (x))) + C

What is the integral of sin^2(ln(x)) ?
The integral of sin^2(ln(x)) is 1/10 (5x-2xsin(2ln(x))-xcos(2ln(x)))+C