What is the integral of 5xcos(7x) ?

The integral of 5xcos(7x) is 5/49 (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 5xcos(7x)

\int 5 x cos (7 x) d x

\frac{5}{49} (7 x sin (7 x) + cos (7 x)) + C

Solution steps

\int 5 x cos (7 x) d x

Take the constant out:

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

= 5 \cdot \int x cos (7 x) d x

Apply u-substitution

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

= 5 \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

= 5 \cdot \frac{1}{49} \cdot \int u cos (u) d u

Apply Integration By Parts

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

= 5 \cdot \frac{1}{49} (u sin (u) - \int sin (u) d u)

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

= 5 \cdot \frac{1}{49} (u sin (u) - (- cos (u)))

Substitute back

u = 7 x

= 5 \cdot \frac{1}{49} (7 x sin (7 x) - (- cos (7 x)))

Simplify

5 \cdot \frac{1}{49} (7 x sin (7 x) - (- cos (7 x))) : \frac{5}{49} (7 x sin (7 x) + cos (7 x))

= \frac{5}{49} (7 x sin (7 x) + cos (7 x))

Add a constant to the solution

= \frac{5}{49} (7 x sin (7 x) + cos (7 x)) + C

\int x^{2} - 8 x d x

\int \frac{( 1 + sin ( t ) ) ^{3}}{cos ( t )} d t

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

\int \frac{2}{x ^{2} + 4 x + 5} d x

\int (\frac{1}{x ^{2}} + \frac{5}{x} - 1) d x

What is the integral of 5xcos(7x) ?
The integral of 5xcos(7x) is 5/49 (7xsin(7x)+cos(7x))+C