What is the integral from 0 to 4pi of x^2sin(2x) ?

The integral from 0 to 4pi of x^2sin(2x) is -8pi^2

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integral from 0 to 4pi of x^2sin(2x)

\int_{0}^{4 π} x^{2} sin (2 x) d x

- 8 π^{2}

Decimal Notation

- 78.95683 \dots

Solution steps

\int_{0}^{4 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}^{4 π}

= [\frac{1}{2} (- x^{2} cos (2 x) - 2 \cdot \int - x cos (2 x) d x)]_{0}^{4 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}^{4 π}

Simplify

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

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

Compute the boundaries:

\frac{- 32 π ^{2} + 1}{4} - \frac{1}{4}

= \frac{- 32 π ^{2} + 1}{4} - \frac{1}{4}

Simplify

= - 8 π^{2}

\int 5 - cos (x) d x

x \to \frac{π}{2} lim (sec^{2} (x))

\frac{\partial}{\partial x} (\frac{x}{( x - 1 ) ^{2}})

\int x \cdot 1 + \frac{1}{4 x} d x

What is the integral from 0 to 4pi of x^2sin(2x) ?
The integral from 0 to 4pi of x^2sin(2x) is -8pi^2