What is the integral of x^2cos(7x) ?

The integral of x^2cos(7x) is 1/343 (49x^2sin(7x)-2(-7xcos(7x)+sin(7x)))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of x^2cos(7x)

\int x^{2} cos (7 x) d x

\frac{1}{343} (49 x^{2} sin (7 x) - 2 (- 7 x cos (7 x) + sin (7 x))) + C

Solution steps

\int x^{2} cos (7 x) d x

Apply u-substitution

: \int \frac{u ^{2} cos ( u )}{343} d u

= \int \frac{u ^{2} cos ( u )}{343} d u

Take the constant out:

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

= \frac{1}{343} \cdot \int u^{2} cos (u) d u

Apply Integration By Parts

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

= \frac{1}{343} (u^{2} sin (u) - \int 2 u sin (u) d u)

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

= \frac{1}{343} (u^{2} sin (u) - 2 (- u cos (u) + sin (u)))

Substitute back

u = 7 x

= \frac{1}{343} ((7 x)^{2} sin (7 x) - 2 (- 7 x cos (7 x) + sin (7 x)))

Simplify

\frac{1}{343} ((7 x)^{2} sin (7 x) - 2 (- 7 x cos (7 x) + sin (7 x))) : \frac{1}{343} (49 x^{2} sin (7 x) - 2 (- 7 x cos (7 x) + sin (7 x)))

= \frac{1}{343} (49 x^{2} sin (7 x) - 2 (- 7 x cos (7 x) + sin (7 x)))

Add a constant to the solution

= \frac{1}{343} (49 x^{2} sin (7 x) - 2 (- 7 x cos (7 x) + sin (7 x))) + C

b cos (t) + t^{2} sin (t)

\int_{- 1}^{1} 3 x^{2} x^{3} + 2 d x

\frac{\partial}{\partial x} (4 x y - x^{4} - y^{4})

y = x^{3}, y = 2 x - x^{2}

What is the integral of x^2cos(7x) ?
The integral of x^2cos(7x) is 1/343 (49x^2sin(7x)-2(-7xcos(7x)+sin(7x)))+C