What is the integral of-8xcos(3x) ?

The integral of-8xcos(3x) is -8/9 (3xsin(3x)+cos(3x))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of-8xcos(3x)

\int - 8 x cos (3 x) d x

- \frac{8}{9} (3 x sin (3 x) + cos (3 x)) + C

Solution steps

\int - 8 x cos (3 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 (3 x) d x

Apply u-substitution

: \int \frac{u cos ( u )}{9} d u

= - 8 \cdot \int \frac{u cos ( u )}{9} d u

Take the constant out:

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

= - 8 \cdot \frac{1}{9} \cdot \int u cos (u) d u

Apply Integration By Parts

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

= - 8 \cdot \frac{1}{9} (u sin (u) - \int sin (u) d u)

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

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

Substitute back

u = 3 x

= - 8 \cdot \frac{1}{9} (3 x sin (3 x) - (- cos (3 x)))

Simplify

- 8 \cdot \frac{1}{9} (3 x sin (3 x) - (- cos (3 x))) : - \frac{8}{9} (3 x sin (3 x) + cos (3 x))

= - \frac{8}{9} (3 x sin (3 x) + cos (3 x))

Add a constant to the solution

= - \frac{8}{9} (3 x sin (3 x) + cos (3 x)) + C

\int e^{- θ} cos (9 θ) d θ

(4 x^{2}) e^{- 10 x}

x \to \frac{π}{2} lim (\frac{tan ( x ) - sec ( x )}{cos ( x )})

x \to 0 lim (7 csc (x) - \frac{7}{x})

y^{^{''}} - (\frac{4}{x}) y^{^{'}} - (\frac{8}{x ^{2}}) y = 0

What is the integral of-8xcos(3x) ?
The integral of-8xcos(3x) is -8/9 (3xsin(3x)+cos(3x))+C