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

The integral from-pi to 0 of x^2sin(nx) is (2-2(-1)^n}{n^3}+\frac{pi^2(-1)^n)/n

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integral from-pi to 0 of x^2sin(nx)

\int_{- π}^{0} x^{2} sin (n x) d x

\frac{2 - 2 ( - 1 ) ^{n}}{n ^{3}} + \frac{π ^{2} ( - 1 ) ^{n}}{n}

Solution steps

\int_{- π}^{0} x^{2} sin (n x) d x

Apply Integration By Parts

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

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

\int - \frac{2}{n} x cos (n x) d x = - \frac{2}{n ^{3}} (n x sin (n x) + cos (n x))

= [- \frac{1}{n} x^{2} cos (n x) - (- \frac{2}{n ^{3}} (n x sin (n x) + cos (n x)))]_{- π}^{0}

Simplify

= [- \frac{1}{n} x^{2} cos (n x) + \frac{2}{n ^{3}} (n x sin (n x) + cos (n x))]_{- π}^{0}

Compute the boundaries:

\frac{2}{n ^{3}} - (- 1)^{n} \frac{2}{n ^{3}} + (- 1)^{n} \frac{π ^{2}}{n}

= \frac{2}{n ^{3}} - (- 1)^{n} \frac{2}{n ^{3}} + (- 1)^{n} \frac{π ^{2}}{n}

Simplify

= \frac{2 - 2 ( - 1 ) ^{n}}{n ^{3}} + \frac{π ^{2} ( - 1 ) ^{n}}{n}

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What is the integral from-pi to 0 of x^2sin(nx) ?
The integral from-pi to 0 of x^2sin(nx) is (2-2(-1)^n}{n^3}+\frac{pi^2(-1)^n)/n