integral of sin^4(x)x

解答

\int sin^{4} (x) x d x

- \frac{1}{4} x sin^{3} (x) cos (x) + \frac{3}{8} x^{2} - \frac{3}{16} x sin (2 x) + \frac{1}{16} sin^{4} (x) - \frac{3 x ^{2}}{16} - \frac{3}{32} cos (2 x) + C

求解步骤

\int sin^{4} (x) x d x

使用分布积分法

= x (- \frac{1}{4} sin^{3} (x) cos (x) + \frac{3}{8} (x - \frac{1}{2} sin (2 x))) - \int - \frac{1}{4} sin^{3} (x) cos (x) + \frac{3}{8} (x - \frac{1}{2} sin (2 x)) d x

\int - \frac{1}{4} sin^{3} (x) cos (x) + \frac{3}{8} (x - \frac{1}{2} sin (2 x)) d x = - \frac{1}{16} sin^{4} (x) + \frac{3}{8} (\frac{x ^{2}}{2} + \frac{1}{4} cos (2 x))

= x (- \frac{1}{4} sin^{3} (x) cos (x) + \frac{3}{8} (x - \frac{1}{2} sin (2 x))) - (- \frac{1}{16} sin^{4} (x) + \frac{3}{8} (\frac{x ^{2}}{2} + \frac{1}{4} cos (2 x)))

化简

x (- \frac{1}{4} sin^{3} (x) cos (x) + \frac{3}{8} (x - \frac{1}{2} sin (2 x))) - (- \frac{1}{16} sin^{4} (x) + \frac{3}{8} (\frac{x ^{2}}{2} + \frac{1}{4} cos (2 x))) : - \frac{1}{4} x sin^{3} (x) cos (x) + \frac{3}{8} x^{2} - \frac{3}{16} x sin (2 x) + \frac{1}{16} sin^{4} (x) - \frac{3 x ^{2}}{16} - \frac{3}{32} cos (2 x)

= - \frac{1}{4} x sin^{3} (x) cos (x) + \frac{3}{8} x^{2} - \frac{3}{16} x sin (2 x) + \frac{1}{16} sin^{4} (x) - \frac{3 x ^{2}}{16} - \frac{3}{32} cos (2 x)

解答补常数

= - \frac{1}{4} x sin^{3} (x) cos (x) + \frac{3}{8} x^{2} - \frac{3}{16} x sin (2 x) + \frac{1}{16} sin^{4} (x) - \frac{3 x ^{2}}{16} - \frac{3}{32} cos (2 x) + C

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