What is the integral of 3xcos(4x) ?

The integral of 3xcos(4x) is 3/16 (4xsin(4x)+cos(4x))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 3xcos(4x)

\int 3 x cos (4 x) d x

\frac{3}{16} (4 x sin (4 x) + cos (4 x)) + C

Solution steps

\int 3 x cos (4 x) d x

Take the constant out:

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

= 3 \cdot \int x cos (4 x) d x

Apply u-substitution

: \int \frac{u cos ( u )}{16} d u

= 3 \cdot \int \frac{u cos ( u )}{16} d u

Take the constant out:

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

= 3 \cdot \frac{1}{16} \cdot \int u cos (u) d u

Apply Integration By Parts

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

= 3 \cdot \frac{1}{16} (u sin (u) - \int sin (u) d u)

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

= 3 \cdot \frac{1}{16} (u sin (u) - (- cos (u)))

Substitute back

u = 4 x

= 3 \cdot \frac{1}{16} (4 x sin (4 x) - (- cos (4 x)))

Simplify

3 \cdot \frac{1}{16} (4 x sin (4 x) - (- cos (4 x))) : \frac{3}{16} (4 x sin (4 x) + cos (4 x))

= \frac{3}{16} (4 x sin (4 x) + cos (4 x))

Add a constant to the solution

= \frac{3}{16} (4 x sin (4 x) + cos (4 x)) + C

\int \frac{3 x ^{2} - 1}{( x - 1 ) x ^{2}} d x

\frac{d}{d x} (e^{c o s (x)})

\int x e^{b x} d x

\int sec^{5} (3 x) d x

\frac{d}{d x} (\frac{2 x y}{( 2 - x ^{2} )})

What is the integral of 3xcos(4x) ?
The integral of 3xcos(4x) is 3/16 (4xsin(4x)+cos(4x))+C