What is the integral from 0 to infinity of sin(9x) ?

The integral from 0 to infinity of sin(9x) is diverges

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to infinity of sin(9x)

\int_{0}^{\infty} sin (9 x) d x

diverges

Solution steps

\int_{0}^{\infty} sin (9 x) d x

Compute the indefinite integral:

\int sin (9 x) d x = - \frac{1}{9} cos (9 x) + C

Compute the boundaries:

\int_{0}^{\infty} sin (9 x) d x =

diverges

= diverges

\frac{d}{d x} (sin (x) - x cos (x))

\int \frac{1}{( 1 - x ) ( 1 - 3 x ) ^{3}} d x

\frac{d}{d x} (\frac{2}{2 x + 3})

\int \frac{6}{x} d x

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

What is the integral from 0 to infinity of sin(9x) ?
The integral from 0 to infinity of sin(9x) is diverges