What is the integral from 0 to pi/5 of t^2sin(5t) ?

The integral from 0 to pi/5 of t^2sin(5t) is (pi^2-4)/(125)

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integral from 0 to pi/5 of t^2sin(5t)

\int_{0}^{\frac{π}{5}} t^{2} sin (5 t) d t

\frac{π ^{2} - 4}{125}

Decimal

0.04695 \dots

Solution steps

\int_{0}^{\frac{π}{5}} t^{2} sin (5 t) d t

Apply Integration By Parts

: [\frac{1}{5} (- t^{2} cos (5 t) - 5 \cdot \int - \frac{2}{5} t cos (5 t) d t)]_{0}^{\frac{π}{5}}

= [\frac{1}{5} (- t^{2} cos (5 t) - 5 \cdot \int - \frac{2}{5} t cos (5 t) d t)]_{0}^{\frac{π}{5}}

\int - \frac{2}{5} t cos (5 t) d t = - \frac{2}{125} (5 t sin (5 t) + cos (5 t))

= [\frac{1}{5} (- t^{2} cos (5 t) - 5 (- \frac{2}{125} (5 t sin (5 t) + cos (5 t))))]_{0}^{\frac{π}{5}}

Simplify

[\frac{1}{5} (- t^{2} cos (5 t) - 5 (- \frac{2}{125} (5 t sin (5 t) + cos (5 t))))]_{0}^{\frac{π}{5}} : [\frac{1}{5} (- t^{2} cos (5 t) + \frac{2}{25} (5 t sin (5 t) + cos (5 t)))]_{0}^{\frac{π}{5}}

= [\frac{1}{5} (- t^{2} cos (5 t) + \frac{2}{25} (5 t sin (5 t) + cos (5 t)))]_{0}^{\frac{π}{5}}

Compute the boundaries:

\frac{π ^{2} - 2}{125} - \frac{2}{125}

= \frac{π ^{2} - 2}{125} - \frac{2}{125}

Simplify

= \frac{π ^{2} - 4}{125}

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What is the integral from 0 to pi/5 of t^2sin(5t) ?
The integral from 0 to pi/5 of t^2sin(5t) is (pi^2-4)/(125)