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

解

\int_{- π}^{π} x^{2} sin^{2} (x) d x

\frac{8 π ^{3} - 12 π}{24}

十進法表記

8.76462 \dots

解答ステップ

\int_{- π}^{π} x^{2} sin^{2} (x) d x

部分積分を適用する

= [\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}

簡素化

[\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}

限度を計算する:

\frac{8 π ^{3} - 12 π}{24}

= \frac{8 π ^{3} - 12 π}{24}