∫sin^2xcos^4xdx求答案
举一反三
- 设$\int_0^\pi {[f(x) + f''(x)]\sin xdx = 5} $，$f(\pi ) = 2$，求$f(0)$=（ ） A: 1 B: 2 C: 3 D: 4
- 已知\( y = \tan x \)，则\( dy \)为（ ）. A: \( \tan xdx \) B: \( \cos xdx \) C: \( {\sec ^2}xdx \) D: \( \sin xdx \)
- 函数y=sin(2x−5)x的导函数为y=2xcos(2x−5)−sin(2x−5)x2y=2xcos(2x−5)−sin(2x−5)x2．
- 证明：sin^4x+cos^4x=1-2sin^2xcos^2x
- 设函数f(x)=sin(x2)+e-2x，则f’(x)等于( )； A: cos(x2)+2e-2x B: 2xcos(x2)-2e-2x C: -2xcos(x2)-e-2x D: 2xcos(x2)+e-2x