What is the integral from-pi to 0 of (t-pi)sin(nt) ?

The integral from-pi to 0 of (t-pi)sin(nt) is (pi-2pi(-1)^n)/n

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integral from-pi to 0 of (t-pi)sin(nt)

\int_{- π}^{0} (t - π) sin (n t) d t

\frac{π - 2 π ( - 1 ) ^{n}}{n}

Solution steps

\int_{- π}^{0} (t - π) sin (n t) d t

Apply Integration By Parts

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

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

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

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

Simplify

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

Compute the boundaries:

\frac{π}{n} - (- 1)^{n} \frac{2 π}{n}

= \frac{π}{n} - (- 1)^{n} \frac{2 π}{n}

Simplify

= \frac{π - 2 π ( - 1 ) ^{n}}{n}

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What is the integral from-pi to 0 of (t-pi)sin(nt) ?
The integral from-pi to 0 of (t-pi)sin(nt) is (pi-2pi(-1)^n)/n