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

The integral from-pi to pi of x^2sin^2(x) is (8pi^3-12pi)/(24)

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

\int_{- π}^{π} x^{2} sin^{2} (x) d x

\frac{8 π ^{3} - 12 π}{24}

Decimal

8.76462 \dots

Solution steps

\int_{- π}^{π} x^{2} sin^{2} (x) d x

Apply Integration By Parts

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

\int \frac{x ^{3} sin ( 2 x )}{3} d x = - \frac{1}{6} x^{3} cos (2 x) + \frac{1}{8} (2 x^{2} sin (2 x) + 2 x cos (2 x) - sin (2 x))

= [\frac{1}{3} (x^{3} sin^{2} (x) - 3 (- \frac{1}{6} x^{3} cos (2 x) + \frac{1}{8} (2 x^{2} sin (2 x) + 2 x cos (2 x) - sin (2 x))))]_{- π}^{π}

Simplify

[\frac{1}{3} (x^{3} sin^{2} (x) - 3 (- \frac{1}{6} x^{3} cos (2 x) + \frac{1}{8} (2 x^{2} sin (2 x) + 2 x cos (2 x) - sin (2 x))))]_{- π}^{π} : [\frac{1}{24} (8 x^{3} sin^{2} (x) + 4 x^{3} cos (2 x) - 6 x^{2} sin (2 x) - 6 x cos (2 x) + 3 sin (2 x))]_{- π}^{π}

= [\frac{1}{24} (8 x^{3} sin^{2} (x) + 4 x^{3} cos (2 x) - 6 x^{2} sin (2 x) - 6 x cos (2 x) + 3 sin (2 x))]_{- π}^{π}

Compute the boundaries:

\frac{8 π ^{3} - 12 π}{24}

= \frac{8 π ^{3} - 12 π}{24}

What is the integral from-pi to pi of x^2sin^2(x) ?
The integral from-pi to pi of x^2sin^2(x) is (8pi^3-12pi)/(24)