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

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

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

\int_{- π}^{0} (π + x) sin (n x) d x

\frac{π ( - 1 + ( - 1 ) ^{n} )}{n} - \frac{π ( - 1 ) ^{n}}{n}

Solution steps

\int_{- π}^{0} (π + x) sin (n x) d x

Expand

(π + x) sin (n x) : π sin (n x) + x sin (n x)

= \int_{- π}^{0} π sin (n x) + x sin (n x) d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{- π}^{0} π sin (n x) d x + \int_{- π}^{0} x sin (n x) d x

\int_{- π}^{0} π sin (n x) d x = \frac{π ( - 1 + ( - 1 ) ^{n} )}{n}

\int_{- π}^{0} x sin (n x) d x = - \frac{π ( - 1 ) ^{n}}{n}

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

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