What is the integral from 0 to 6pi of x^2sin(2x) ?

The integral from 0 to 6pi of x^2sin(2x) is -18pi^2

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integral from 0 to 6pi of x^2sin(2x)

\int_{0}^{6 π} x^{2} sin (2 x) d x

- 18 π^{2}

Decimal Notation

- 177.65287 \dots

Solution steps

\int_{0}^{6 π} x^{2} sin (2 x) d x

Apply Integration By Parts

: [\frac{1}{2} (- x^{2} cos (2 x) - 2 \cdot \int - x cos (2 x) d x)]_{0}^{6 π}

= [\frac{1}{2} (- x^{2} cos (2 x) - 2 \cdot \int - x cos (2 x) d x)]_{0}^{6 π}

\int - x cos (2 x) d x = - \frac{1}{4} (2 x sin (2 x) + cos (2 x))

= [\frac{1}{2} (- x^{2} cos (2 x) - 2 (- \frac{1}{4} (2 x sin (2 x) + cos (2 x))))]_{0}^{6 π}

Simplify

[\frac{1}{2} (- x^{2} cos (2 x) - 2 (- \frac{1}{4} (2 x sin (2 x) + cos (2 x))))]_{0}^{6 π} : [\frac{1}{4} (- 2 x^{2} cos (2 x) + 2 x sin (2 x) + cos (2 x))]_{0}^{6 π}

= [\frac{1}{4} (- 2 x^{2} cos (2 x) + 2 x sin (2 x) + cos (2 x))]_{0}^{6 π}

Compute the boundaries:

\frac{- 72 π ^{2} + 1}{4} - \frac{1}{4}

= \frac{- 72 π ^{2} + 1}{4} - \frac{1}{4}

Simplify

= - 18 π^{2}

ln (x^{2} - 4 x + 1)

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

\frac{d y}{d x}, y^{3} - 2 x y + 7 = 3 x + 1

\int 5 x \cdot cos (8 x) d x

What is the integral from 0 to 6pi of x^2sin(2x) ?
The integral from 0 to 6pi of x^2sin(2x) is -18pi^2