What is the integral from 0 to pi of t^2sin(nt) ?

The integral from 0 to pi of t^2sin(nt) is (2(-1)^n-pi^2(-1)^nn^2-2)/(n^3)

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integral from 0 to pi of t^2sin(nt)

\int_{0}^{π} t^{2} sin (n t) d t

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

Solution steps

\int_{0}^{π} t^{2} sin (n t) d t

Apply Integration By Parts

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

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

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

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

Simplify

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

= [\frac{1}{n} (- t^{2} cos (n t) + \frac{2}{n ^{2}} (n t sin (n t) + cos (n t)))]_{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 t^2sin(nt) ?
The integral from 0 to pi of t^2sin(nt) is (2(-1)^n-pi^2(-1)^nn^2-2)/(n^3)