What is the integral of 3xcos(2x) ?

The integral of 3xcos(2x) is 3/4 (2xsin(2x)+cos(2x))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 3xcos(2x)

\int 3 x cos (2 x) d x

\frac{3}{4} (2 x sin (2 x) + cos (2 x)) + C

Solution steps

\int 3 x cos (2 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 (2 x) d x

Apply u-substitution

: \int \frac{u cos ( u )}{4} d u

= 3 \cdot \int \frac{u cos ( u )}{4} d u

Take the constant out:

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

= 3 \cdot \frac{1}{4} \cdot \int u cos (u) d u

Apply Integration By Parts

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

= 3 \cdot \frac{1}{4} (u sin (u) - \int sin (u) d u)

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

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

Substitute back

u = 2 x

= 3 \cdot \frac{1}{4} (2 x sin (2 x) - (- cos (2 x)))

Simplify

3 \cdot \frac{1}{4} (2 x sin (2 x) - (- cos (2 x))) : \frac{3}{4} (2 x sin (2 x) + cos (2 x))

= \frac{3}{4} (2 x sin (2 x) + cos (2 x))

Add a constant to the solution

= \frac{3}{4} (2 x sin (2 x) + cos (2 x)) + C

\int 3 5 - 3 x^{2} (- 6 x) d x

\int_{0}^{x b} \frac{( 1 - 0.25 x ) ^{3}}{( 1 - \frac{1}{2} x ) ( 1 - x ) ^{2}} d x

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

x \to - 2 lim (x^{2} + 2 t - 1)

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

What is the integral of 3xcos(2x) ?
The integral of 3xcos(2x) is 3/4 (2xsin(2x)+cos(2x))+C