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

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

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

\int_{0}^{8 π} t^{2} sin (2 t) d t

- 32 π^{2}

Decimal

- 315.82734 \dots

Solution steps

\int_{0}^{8 π} t^{2} sin (2 t) d t

Apply Integration By Parts

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

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

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

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

Simplify

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

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

\int a x + b d x

\int \frac{e ^{x} ( x - 1 ) - e ^{x}}{( x - 1 ) ^{2}} d x

tan (\frac{1}{x})

\frac{1}{2 - x}, 1

\frac{d}{d x} (\frac{3 x ^{3} + x + 2}{5 x ^{2} + 1})

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