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

The integral from 0 to 8pi of x^2sin(2x) is -32pi^2

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integral from 0 to 8pi of x^2sin(2x)

\int_{0}^{8 π} x^{2} sin (2 x) d x

- 32 π^{2}

Decimal

- 315.82734 \dots

Solution steps

\int_{0}^{8 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}^{8 π}

= [\frac{1}{2} (- x^{2} cos (2 x) - 2 \cdot \int - x cos (2 x) d x)]_{0}^{8 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}^{8 π}

Simplify

[\frac{1}{2} (- x^{2} cos (2 x) - 2 (- \frac{1}{4} (2 x sin (2 x) + cos (2 x))))]_{0}^{8 π} : [\frac{1}{4} (- 2 x^{2} cos (2 x) + 2 x sin (2 x) + cos (2 x))]_{0}^{8 π}

= [\frac{1}{4} (- 2 x^{2} cos (2 x) + 2 x sin (2 x) + cos (2 x))]_{0}^{8 π}

Compute the boundaries:

\frac{- 128 π ^{2} + 1}{4} - \frac{1}{4}

= \frac{- 128 π ^{2} + 1}{4} - \frac{1}{4}

Simplify

= - 32 π^{2}

sin (tan (ln (x^{2} + 5)))

f (x) = - 15 x^{2} - 14 x + 8

t^{3} e^{- 0.1 t}

\int_{1}^{\infty} \frac{1}{e ^{x}} d x

\int sin (3 x) sin (5 x) d x

What is the integral from 0 to 8pi of x^2sin(2x) ?
The integral from 0 to 8pi of x^2sin(2x) is -32pi^2