What is the integral of 3x^2cos(6x) ?

The integral of 3x^2cos(6x) is 1/36 (18x^2sin(6x)-sin(6x)+6xcos(6x))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 3x^2cos(6x)

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

\frac{1}{36} (18 x^{2} sin (6 x) - sin (6 x) + 6 x cos (6 x)) + C

Solution steps

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

Apply u-substitution

: \int \frac{u ^{2} cos ( u )}{216} d u

= 3 \cdot \int \frac{u ^{2} cos ( u )}{216} d u

Take the constant out:

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

= 3 \cdot \frac{1}{216} \cdot \int u^{2} cos (u) d u

Apply Integration By Parts

: u^{2} sin (u) - \int 2 u sin (u) d u

= 3 \cdot \frac{1}{216} (u^{2} sin (u) - \int 2 u sin (u) d u)

\int 2 u sin (u) d u = 2 (- u cos (u) + sin (u))

= 3 \cdot \frac{1}{216} (u^{2} sin (u) - 2 (- u cos (u) + sin (u)))

Substitute back

u = 6 x

= 3 \cdot \frac{1}{216} ((6 x)^{2} sin (6 x) - 2 (- 6 x cos (6 x) + sin (6 x)))

Simplify

3 \cdot \frac{1}{216} ((6 x)^{2} sin (6 x) - 2 (- 6 x cos (6 x) + sin (6 x))) : \frac{1}{36} (18 x^{2} sin (6 x) - sin (6 x) + 6 x cos (6 x))

= \frac{1}{36} (18 x^{2} sin (6 x) - sin (6 x) + 6 x cos (6 x))

Add a constant to the solution

= \frac{1}{36} (18 x^{2} sin (6 x) - sin (6 x) + 6 x cos (6 x)) + C

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

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

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

2^{z} lo g_{2} (z)

y^{^{'}} + (\frac{3 y}{( 300 - t )}) = 4

What is the integral of 3x^2cos(6x) ?
The integral of 3x^2cos(6x) is 1/36 (18x^2sin(6x)-sin(6x)+6xcos(6x))+C