integral of t^2cos(2t)

解答

\int t^{2} cos (2 t) d t

\frac{1}{2} t^{2} sin (2 t) + \frac{1}{2} t cos (2 t) - \frac{1}{4} sin (2 t) + C

求解步骤

\int t^{2} cos (2 t) d t

使用分布积分法

= \frac{1}{2} t^{2} sin (2 t) - \int t sin (2 t) d t

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

= \frac{1}{2} t^{2} sin (2 t) - (- \frac{1}{2} t cos (2 t) + \frac{1}{4} sin (2 t))

化简

= \frac{1}{2} t^{2} sin (2 t) + \frac{1}{2} t cos (2 t) - \frac{1}{4} sin (2 t)

解答补常数

= \frac{1}{2} t^{2} sin (2 t) + \frac{1}{2} t cos (2 t) - \frac{1}{4} sin (2 t) + C

\frac{d}{d x} ((- x^{3} - x^{2} + x) cos (x))

in v erse l a pl a ce \frac{1}{p ( p + 1 )}

t an g e n t y = tan (18 x^{2}), at x = π

y^{^{''}} + 9 y = csc^{2} (3 t)

in v erse l a pl a ce \frac{5}{s ^{2} + 4}