∫(π/2 0)sin(x+π/2)dx=?A.-2 B.-1 C.1 D.2
来源:学生作业帮助网 编辑:六六作业网 时间:2024/12/19 23:01:29
∫(π/20)sin(x+π/2)dx=?A.-2B.-1C.1D.2∫(π/20)sin(x+π/2)dx=?A.-2B.-1C.1D.2∫(π/20)sin(x+π/2)dx=?A.-2B.-1C
∫(π/2 0)sin(x+π/2)dx=?A.-2 B.-1 C.1 D.2
∫(π/2 0)sin(x+π/2)dx=?A.-2 B.-1 C.1 D.2
∫(π/2 0)sin(x+π/2)dx=?A.-2 B.-1 C.1 D.2
∫ sin(x+π/2) dx
=∫ cosx dx
= sinx
那么代入上下限π/2和0
得到
原积分= 1- 0=1
选择C
求定积分 ∫[0,π]sin 2x dx
∫(0→2π) |sin x | dx
求∫e^sin(πx/2)dx的积分....
证明∫(0,π/2)sin^m x dx=∫(0,π/2)cos^m x dx
证明∫( 0,π/2 ) (f sin x/(f sin x+f cos x) dx=π /4
∫e^sin x/(e^sin x+e^cos x)dx在0~π/2上的积分
求定积分∫(0到π/2)[cos(x/2)-sin(x/2)]^2dx
求定积分 ∫ (π/2 ,0) x sin x dx
∫x/sin^2(x) dx
∫sin(x) cos^2(x)dx
∫[0,π/2]sin^2X/2dx求定积分
∫(0,π/2)sinx/(3+sin^2x)dx求定积分
求定积分∫(0,π/2) sin²(x/2)dx
- ∫(0->π/2) (1+cosx)²sin³x(1+2cosx)dx
∫[0,π]sin^2X/2dx求定积分
求定积分∫(上限为π/2.下限为0)|1/2-sin x| dx
求定积分 ∫ (π→0) √(1+sin 2x ) dx
∫sin^2x/(1+sin^2x )dx求解,