What is the integral from 0 to pi of x^2sin(3x) ?

The integral from 0 to pi of x^2sin(3x) is (9pi^2-4)/(27)

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integral from 0 to pi of x^2sin(3x)

\int_{0}^{π} x^{2} sin (3 x) d x

\frac{9 π ^{2} - 4}{27}

Decimal Notation

3.14171 \dots

Solution steps

\int_{0}^{π} x^{2} sin (3 x) d x

Apply Integration By Parts

: [\frac{1}{3} (- x^{2} cos (3 x) - 3 \cdot \int - \frac{2}{3} x cos (3 x) d x)]_{0}^{π}

= [\frac{1}{3} (- x^{2} cos (3 x) - 3 \cdot \int - \frac{2}{3} x cos (3 x) d x)]_{0}^{π}

\int - \frac{2}{3} x cos (3 x) d x = - \frac{2}{27} (3 x sin (3 x) + cos (3 x))

= [\frac{1}{3} (- x^{2} cos (3 x) - 3 (- \frac{2}{27} (3 x sin (3 x) + cos (3 x))))]_{0}^{π}

Simplify

[\frac{1}{3} (- x^{2} cos (3 x) - 3 (- \frac{2}{27} (3 x sin (3 x) + cos (3 x))))]_{0}^{π} : [\frac{1}{3} (- x^{2} cos (3 x) + \frac{2}{9} (3 x sin (3 x) + cos (3 x)))]_{0}^{π}

= [\frac{1}{3} (- x^{2} cos (3 x) + \frac{2}{9} (3 x sin (3 x) + cos (3 x)))]_{0}^{π}

Compute the boundaries:

\frac{9 π ^{2} - 2}{27} - \frac{2}{27}

= \frac{9 π ^{2} - 2}{27} - \frac{2}{27}

Simplify

= \frac{9 π ^{2} - 4}{27}

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What is the integral from 0 to pi of x^2sin(3x) ?
The integral from 0 to pi of x^2sin(3x) is (9pi^2-4)/(27)