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

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

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

\int_{0}^{2 π} t^{2} sin (2 t) d t

- 2 π^{2}

Decimal Notation

- 19.73920 \dots

Solution steps

\int_{0}^{2 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}^{2 π}

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

Simplify

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

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

Compute the boundaries:

\frac{- 8 π ^{2} + 1}{4} - \frac{1}{4}

= \frac{- 8 π ^{2} + 1}{4} - \frac{1}{4}

Simplify

= - 2 π^{2}

f (x) = \frac{32}{x ^{3}}

x + 2, 1, 3

\frac{d}{d x} (x ln (x) cos (x))

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

\int + \frac{1}{25 e ^{- 4 x} + e ^{4 x}} d x

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