What is the integral of x^2cos(21x) ?

The integral of x^2cos(21x) is 1/9261 (441x^2sin(21x)-2(-21xcos(21x)+sin(21x)))+C

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integral of x^2cos(21x)

\int x^{2} cos (21 x) d x

\frac{1}{9261} (441 x^{2} sin (21 x) - 2 (- 21 x cos (21 x) + sin (21 x))) + C

Solution steps

\int x^{2} cos (21 x) d x

Apply u-substitution

: \int \frac{u ^{2} cos ( u )}{9261} d u

= \int \frac{u ^{2} cos ( u )}{9261} d u

Take the constant out:

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

= \frac{1}{9261} \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}{9261} (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}{9261} (u^{2} sin (u) - 2 (- u cos (u) + sin (u)))

Substitute back

u = 21 x

= \frac{1}{9261} ((21 x)^{2} sin (21 x) - 2 (- 21 x cos (21 x) + sin (21 x)))

Simplify

\frac{1}{9261} ((21 x)^{2} sin (21 x) - 2 (- 21 x cos (21 x) + sin (21 x))) : \frac{1}{9261} (441 x^{2} sin (21 x) - 2 (- 21 x cos (21 x) + sin (21 x)))

= \frac{1}{9261} (441 x^{2} sin (21 x) - 2 (- 21 x cos (21 x) + sin (21 x)))

Add a constant to the solution

= \frac{1}{9261} (441 x^{2} sin (21 x) - 2 (- 21 x cos (21 x) + sin (21 x))) + C

\int_{0}^{5} ∣ 3 - x ∣ d x

\frac{\partial}{\partial x} (y \cdot r^{2})

\int \frac{5 ln ( x )}{x} d x

y^{^{'}} = (y - 1) (y - 1)

\int_{1}^{5} 2 d y

What is the integral of x^2cos(21x) ?
The integral of x^2cos(21x) is 1/9261 (441x^2sin(21x)-2(-21xcos(21x)+sin(21x)))+C