What is the integral of 3xcos(9x) ?

The integral of 3xcos(9x) is 1/27 (9xsin(9x)+cos(9x))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 3xcos(9x)

\int 3 x cos (9 x) d x

\frac{1}{27} (9 x sin (9 x) + cos (9 x)) + C

Solution steps

\int 3 x cos (9 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 (9 x) d x

Apply u-substitution

: \int \frac{u cos ( u )}{81} d u

= 3 \cdot \int \frac{u cos ( u )}{81} d u

Take the constant out:

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

= 3 \cdot \frac{1}{81} \cdot \int u cos (u) d u

Apply Integration By Parts

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

= 3 \cdot \frac{1}{81} (u sin (u) - \int sin (u) d u)

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

= 3 \cdot \frac{1}{81} (u sin (u) - (- cos (u)))

Substitute back

u = 9 x

= 3 \cdot \frac{1}{81} (9 x sin (9 x) - (- cos (9 x)))

Simplify

3 \cdot \frac{1}{81} (9 x sin (9 x) - (- cos (9 x))) : \frac{1}{27} (9 x sin (9 x) + cos (9 x))

= \frac{1}{27} (9 x sin (9 x) + cos (9 x))

Add a constant to the solution

= \frac{1}{27} (9 x sin (9 x) + cos (9 x)) + C

f (x) = x^{3} - 2 x + 1

800 x^{2} - 0.3 x

\frac{d}{d x} (lo g_{3} (sec (x)))

f (x) = \frac{x}{1 + x ^{2}}, (0, 0)

\frac{d}{d x} (\frac{3 - sec ( x )}{tan ( x )})

What is the integral of 3xcos(9x) ?
The integral of 3xcos(9x) is 1/27 (9xsin(9x)+cos(9x))+C