What is the integral of-2xcos(8x) ?

The integral of-2xcos(8x) is -1/32 (8xsin(8x)+cos(8x))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of-2xcos(8x)

\int - 2 x cos (8 x) d x

- \frac{1}{32} (8 x sin (8 x) + cos (8 x)) + C

Solution steps

\int - 2 x cos (8 x) d x

Take the constant out:

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

= - 2 \cdot \int x cos (8 x) d x

Apply u-substitution

= - 2 \cdot \int \frac{u cos ( u )}{64} d u

Take the constant out:

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

= - 2 \cdot \frac{1}{64} \cdot \int u cos (u) d u

Apply Integration By Parts

= - 2 \cdot \frac{1}{64} (u sin (u) - \int sin (u) d u)

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

= - 2 \cdot \frac{1}{64} (u sin (u) - (- cos (u)))

Substitute back

u = 8 x

= - 2 \cdot \frac{1}{64} (8 x sin (8 x) - (- cos (8 x)))

Simplify

- 2 \cdot \frac{1}{64} (8 x sin (8 x) - (- cos (8 x))) : - \frac{1}{32} (8 x sin (8 x) + cos (8 x))

= - \frac{1}{32} (8 x sin (8 x) + cos (8 x))

Add a constant to the solution

= - \frac{1}{32} (8 x sin (8 x) + cos (8 x)) + C

in v erse l a pl a ce \frac{5}{s ( 4 s ^{2} + 4 s + 1 )}

x \to 3 lim (- x 6 x - 8)

\int_{- \infty}^{1} 3^{x} d x

\frac{d}{d x} (3 π^{8})

t an g e n t f (x) = - 4 x^{\frac{1}{2}} - 10, at x = 9

What is the integral of-2xcos(8x) ?
The integral of-2xcos(8x) is -1/32 (8xsin(8x)+cos(8x))+C