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

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

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

\int_{0}^{6 π} t^{2} sin (2 t) d t

- 18 π^{2}

Decimal Notation

- 177.65287 \dots

Solution steps

\int_{0}^{6 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}^{6 π}

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

Simplify

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

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

\int 6 (x - 12)^{5} d x

f (x) = x^{2} - 6, at x = 3

6 e^{z}

\int \frac{1}{x ( x - 7 )} d x

\frac{d}{d y} (\frac{1}{1 + y ^{2}})

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