What is the integral from-pi to 0 of (x-pi)sin(nx) ?

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

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integral from-pi to 0 of (x-pi)sin(nx)

\int_{- π}^{0} (x - π) sin (n x) d x

\frac{π - 2 π ( - 1 ) ^{n}}{n}

Solution steps

\int_{- π}^{0} (x - π) sin (n x) d x

Apply Integration By Parts

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

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

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

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

Simplify

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

\int_{4}^{8} x^{2} - 16 d x

\int_{- 1}^{1} 7 t^{6} d t

\int_{0}^{1} t^{2} + 2 d t

\int_{y}^{y} 1 d x

\int_{0}^{2} \frac{1}{6} x d x

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