integral from pi/2 to pi of xsin^2(x)

解

\int_{\frac{π}{2}}^{π} x sin^{2} (x) d x

\frac{3 π ^{2} - 4}{16}

十進法表記

1.60055 \dots

解答ステップ

\int_{\frac{π}{2}}^{π} x sin^{2} (x) d x

部分積分を適用する

= [\frac{1}{4} (x (2 x - sin (2 x)) - 4 \cdot \int \frac{1}{2} (x - \frac{1}{2} sin (2 x)) d x)]_{\frac{π}{2}}^{π}

\int \frac{1}{2} (x - \frac{1}{2} sin (2 x)) d x = \frac{1}{2} (\frac{x ^{2}}{2} + \frac{1}{4} cos (2 x))

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

簡素化

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

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

限度を計算する:

\frac{3 π ^{2} - 4}{16}

= \frac{3 π ^{2} - 4}{16}