integral of x^2sin^2(2x)

解答

\int x^{2} sin^{2} (2 x) d x

\frac{1}{3} x^{3} sin^{2} (2 x) + \frac{1}{6} x^{3} cos (4 x) - \frac{1}{8} x^{2} sin (4 x) - \frac{1}{16} x cos (4 x) + \frac{1}{64} sin (4 x) + C

求解步骤

\int x^{2} sin^{2} (2 x) d x

使用分布积分法

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

\int \frac{2}{3} x^{3} sin (4 x) d x = - \frac{1}{6} x^{3} cos (4 x) + \frac{1}{64} (8 x^{2} sin (4 x) + 4 x cos (4 x) - sin (4 x))

= \frac{1}{3} x^{3} sin^{2} (2 x) - (- \frac{1}{6} x^{3} cos (4 x) + \frac{1}{64} (8 x^{2} sin (4 x) + 4 x cos (4 x) - sin (4 x)))

化简

\frac{1}{3} x^{3} sin^{2} (2 x) - (- \frac{1}{6} x^{3} cos (4 x) + \frac{1}{64} (8 x^{2} sin (4 x) + 4 x cos (4 x) - sin (4 x))) : \frac{1}{3} x^{3} sin^{2} (2 x) + \frac{1}{6} x^{3} cos (4 x) - \frac{1}{8} x^{2} sin (4 x) - \frac{1}{16} x cos (4 x) + \frac{1}{64} sin (4 x)

= \frac{1}{3} x^{3} sin^{2} (2 x) + \frac{1}{6} x^{3} cos (4 x) - \frac{1}{8} x^{2} sin (4 x) - \frac{1}{16} x cos (4 x) + \frac{1}{64} sin (4 x)

解答补常数

= \frac{1}{3} x^{3} sin^{2} (2 x) + \frac{1}{6} x^{3} cos (4 x) - \frac{1}{8} x^{2} sin (4 x) - \frac{1}{16} x cos (4 x) + \frac{1}{64} sin (4 x) + C

\int \frac{2 sin ( 2 x )}{1 + cos ( x )} d x

\int - \frac{1}{3} d x

\int 2 \cdot e^{x} d x

\int x y e^{x^{2} y} d x

\int \frac{x}{64 + x ^{4}} d x