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

The integral from 0 to pi of x^2*sin(nx) is (2(-1)^n-pi^2(-1)^nn^2-2)/(n^3)

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

\int_{0}^{π} x^{2} \cdot sin (n x) d x

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

Solution steps

\int_{0}^{π} x^{2} sin (n x) d x

Apply Integration By Parts

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

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

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

Simplify

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

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

Compute the boundaries:

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

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

Simplify

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

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What is the integral from 0 to pi of x^2*sin(nx) ?
The integral from 0 to pi of x^2*sin(nx) is (2(-1)^n-pi^2(-1)^nn^2-2)/(n^3)