What is the integral of 5x^2cos(3x) ?

The integral of 5x^2cos(3x) is 5/27 (9x^2sin(3x)-2(-3xcos(3x)+sin(3x)))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 5x^2cos(3x)

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

\frac{5}{27} (9 x^{2} sin (3 x) - 2 (- 3 x cos (3 x) + sin (3 x))) + C

Solution steps

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

Apply u-substitution

: \int \frac{u ^{2} cos ( u )}{27} d u

= 5 \cdot \int \frac{u ^{2} cos ( u )}{27} d u

Take the constant out:

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

= 5 \cdot \frac{1}{27} \cdot \int u^{2} cos (u) d u

Apply Integration By Parts

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

= 5 \cdot \frac{1}{27} (u^{2} sin (u) - \int 2 u sin (u) d u)

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

= 5 \cdot \frac{1}{27} (u^{2} sin (u) - 2 (- u cos (u) + sin (u)))

Substitute back

u = 3 x

= 5 \cdot \frac{1}{27} ((3 x)^{2} sin (3 x) - 2 (- 3 x cos (3 x) + sin (3 x)))

Simplify

5 \cdot \frac{1}{27} ((3 x)^{2} sin (3 x) - 2 (- 3 x cos (3 x) + sin (3 x))) : \frac{5}{27} (9 x^{2} sin (3 x) - 2 (- 3 x cos (3 x) + sin (3 x)))

= \frac{5}{27} (9 x^{2} sin (3 x) - 2 (- 3 x cos (3 x) + sin (3 x)))

Add a constant to the solution

= \frac{5}{27} (9 x^{2} sin (3 x) - 2 (- 3 x cos (3 x) + sin (3 x))) + C

\int 7 x^{8} - 0^{2} (2 x^{3}) d x

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

\int \frac{tan ^{3} ( x )}{sec ^{2} ( x )} d x

\int (x - 1) (x - 2) d x

What is the integral of 5x^2cos(3x) ?
The integral of 5x^2cos(3x) is 5/27 (9x^2sin(3x)-2(-3xcos(3x)+sin(3x)))+C