What is the integral from-pi to pi of tsin(nt) ?

The integral from-pi to pi of tsin(nt) is -(2pi(-1)^n)/n

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integral from-pi to pi of tsin(nt)

\int_{- π}^{π} t sin (n t) d t

- \frac{2 π ( - 1 ) ^{n}}{n}

Solution steps

\int_{- π}^{π} t sin (n t) d t

Apply Integration By Parts

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

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

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

Simplify

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

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

Compute the boundaries:

- 2 (- 1)^{n} \frac{π}{n}

= - 2 (- 1)^{n} \frac{π}{n}

Simplify

= - \frac{2 π ( - 1 ) ^{n}}{n}

What is the integral from-pi to pi of tsin(nt) ?
The integral from-pi to pi of tsin(nt) is -(2pi(-1)^n)/n