What is the integral of 8xcos(9x) ?

The integral of 8xcos(9x) is 8/81 (9xsin(9x)+cos(9x))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 8xcos(9x)

\int 8 x cos (9 x) d x

\frac{8}{81} (9 x sin (9 x) + cos (9 x)) + C

Solution steps

\int 8 x cos (9 x) d x

Take the constant out:

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

= 8 \cdot \int x cos (9 x) d x

Apply u-substitution

: \int \frac{u cos ( u )}{81} d u

= 8 \cdot \int \frac{u cos ( u )}{81} d u

Take the constant out:

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

= 8 \cdot \frac{1}{81} \cdot \int u cos (u) d u

Apply Integration By Parts

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

= 8 \cdot \frac{1}{81} (u sin (u) - \int sin (u) d u)

\int sin (u) d u = - cos (u)

= 8 \cdot \frac{1}{81} (u sin (u) - (- cos (u)))

Substitute back

u = 9 x

= 8 \cdot \frac{1}{81} (9 x sin (9 x) - (- cos (9 x)))

Simplify

8 \cdot \frac{1}{81} (9 x sin (9 x) - (- cos (9 x))) : \frac{8}{81} (9 x sin (9 x) + cos (9 x))

= \frac{8}{81} (9 x sin (9 x) + cos (9 x))

Add a constant to the solution

= \frac{8}{81} (9 x sin (9 x) + cos (9 x)) + C

\frac{d}{d θ} (cos (θ))

y = \frac{2 x + 1}{x + 2}

\int e^{- s t} sin (2 t) d t

\frac{d}{d x} (tanh (e^{x}))

\frac{10}{s \cdot ( s + 5 ) ^{2}}

What is the integral of 8xcos(9x) ?
The integral of 8xcos(9x) is 8/81 (9xsin(9x)+cos(9x))+C