What is the integral of 8xcos(7x) ?

The integral of 8xcos(7x) is 8/49 (7xsin(7x)+cos(7x))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 8xcos(7x)

\int 8 x cos (7 x) d x

\frac{8}{49} (7 x sin (7 x) + cos (7 x)) + C

Solution steps

\int 8 x cos (7 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 (7 x) d x

Apply u-substitution

: \int \frac{u cos ( u )}{49} d u

= 8 \cdot \int \frac{u cos ( u )}{49} d u

Take the constant out:

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

= 8 \cdot \frac{1}{49} \cdot \int u cos (u) d u

Apply Integration By Parts

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

= 8 \cdot \frac{1}{49} (u sin (u) - \int sin (u) d u)

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

= 8 \cdot \frac{1}{49} (u sin (u) - (- cos (u)))

Substitute back

u = 7 x

= 8 \cdot \frac{1}{49} (7 x sin (7 x) - (- cos (7 x)))

Simplify

8 \cdot \frac{1}{49} (7 x sin (7 x) - (- cos (7 x))) : \frac{8}{49} (7 x sin (7 x) + cos (7 x))

= \frac{8}{49} (7 x sin (7 x) + cos (7 x))

Add a constant to the solution

= \frac{8}{49} (7 x sin (7 x) + cos (7 x)) + C

\int \frac{x ^{4} - 3 x ^{\frac{1}{2}} + 2}{x ^{\frac{3}{2}}} d x

f (x) = \frac{x ^{3}}{5 - x ^{2}}

\int_{0}^{2 π} sin (x) cos^{2} (x) d x

\int sec (y) (tan (y) - sec (y)) d y

\frac{d}{d x} (e^{(x + 1)^{2}})

What is the integral of 8xcos(7x) ?
The integral of 8xcos(7x) is 8/49 (7xsin(7x)+cos(7x))+C