What is the integral from 0 to pi/3 of t^2sin(3t) ?

The integral from 0 to pi/3 of t^2sin(3t) is (pi^2-4)/(27)

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integral from 0 to pi/3 of t^2sin(3t)

\int_{0}^{\frac{π}{3}} t^{2} sin (3 t) d t

\frac{π ^{2} - 4}{27}

Decimal

0.21739 \dots

Solution steps

\int_{0}^{\frac{π}{3}} t^{2} sin (3 t) d t

Apply Integration By Parts

: [\frac{1}{3} (- t^{2} cos (3 t) - 3 \cdot \int - \frac{2}{3} t cos (3 t) d t)]_{0}^{\frac{π}{3}}

= [\frac{1}{3} (- t^{2} cos (3 t) - 3 \cdot \int - \frac{2}{3} t cos (3 t) d t)]_{0}^{\frac{π}{3}}

\int - \frac{2}{3} t cos (3 t) d t = - \frac{2}{27} (3 t sin (3 t) + cos (3 t))

= [\frac{1}{3} (- t^{2} cos (3 t) - 3 (- \frac{2}{27} (3 t sin (3 t) + cos (3 t))))]_{0}^{\frac{π}{3}}

Simplify

[\frac{1}{3} (- t^{2} cos (3 t) - 3 (- \frac{2}{27} (3 t sin (3 t) + cos (3 t))))]_{0}^{\frac{π}{3}} : [\frac{1}{3} (- t^{2} cos (3 t) + \frac{2}{9} (3 t sin (3 t) + cos (3 t)))]_{0}^{\frac{π}{3}}

= [\frac{1}{3} (- t^{2} cos (3 t) + \frac{2}{9} (3 t sin (3 t) + cos (3 t)))]_{0}^{\frac{π}{3}}

Compute the boundaries:

\frac{π ^{2} - 2}{27} - \frac{2}{27}

= \frac{π ^{2} - 2}{27} - \frac{2}{27}

Simplify

= \frac{π ^{2} - 4}{27}

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What is the integral from 0 to pi/3 of t^2sin(3t) ?
The integral from 0 to pi/3 of t^2sin(3t) is (pi^2-4)/(27)