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

The integral from 0 to pi of tsin(2t) is -pi/2

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integral from 0 to pi of tsin(2t)

\int_{0}^{π} t sin (2 t) d t

- \frac{π}{2}

Solution steps

\int_{0}^{π} t sin (2 t) d t

Apply Integration By Parts

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

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

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

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

Simplify

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

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

Compute the boundaries:

- \frac{π}{2}

= - \frac{π}{2}

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