证明∫(0,π/2)sin^m x dx=∫(0,π/2)cos^m x dx
来源:学生作业帮助网 编辑:六六作业网 时间:2024/12/20 08:12:25
证明∫(0,π/2)sin^mxdx=∫(0,π/2)cos^mxdx证明∫(0,π/2)sin^mxdx=∫(0,π/2)cos^mxdx证明∫(0,π/2)sin^mxdx=∫(0,π/2)cos
证明∫(0,π/2)sin^m x dx=∫(0,π/2)cos^m x dx
证明∫(0,π/2)sin^m x dx=∫(0,π/2)cos^m x dx
证明∫(0,π/2)sin^m x dx=∫(0,π/2)cos^m x dx
令x = π/2 - u => dx = - du
∫(0~π/2) sin^m(x) dx
= ∫(π/2~0) [sin(π/2 - u)]^m (- du)
= -∫(π/2~0) [cos(u)]^m du
= ∫(0~π/2) cos^m(u) du
= ∫(0~π/2) cos^m(x) dx,得证
证明∫(0,π/2)sin^m x dx=∫(0,π/2)cos^m x dx
证明∫(0,π/2)sin^m x cos^m x dx=1/2^m∫(0,π/2)cos^m xdx
证明∫(0,π/2)sin^m x cos^m x dx=1/2^m∫(0,π/2)cos^m xdx
证明∫( 0,π/2 ) (f sin x/(f sin x+f cos x) dx=π /4
证明∫sin(x^2)dx=0.5√(π/2),积分区间为0到正无穷.
证明∫sin(x^2) dx >0 积分区间为0到√(pi/2)
求定积分 ∫[0,π]sin 2x dx
∫(0→2π) |sin x | dx
证明∫xf(sin x)dx=π/2∫f(sin x)dx 积分区间都是0到π我自己反复用代换做过了但是一直卡住。
∫ sin(mx)cos(nx) 怎么等于1/2 ∫ [sin(m+n)x + sin(m-n)x] dx
∫x/sin^2(x) dx
∫sin(x) cos^2(x)dx
∫e^sin x/(e^sin x+e^cos x)dx在0~π/2上的积分
高数 积分 证明∫(0~+∞)e^-x x^m dx=m!
求∫e^sin(πx/2)dx的积分....
∫sin^2x/(1+sin^2x )dx求解,
∫sin(6x)sin(2x)dx=?
∫(0,PI/2) sin^4 (x) dx