求∫cos(3x+1)dx.
来源:学生作业帮助网 编辑:作业帮 时间:2024/11/05 14:39:27
求∫cos(3x+1)dx.
求∫cos(3x+1)dx.
求∫cos(3x+1)dx.
∫cos(3x+1)dx=(1/3)∫cos(3x+1)d3x=(1/3)∫cos(3x+1)d(3x+1)=(1/3)∫d(sin(3x+1))=(1/3)sin(3x+1)+C
求∫cos(3x+1)dx.
求∫[0-->π](1-cos³x )dx
求∫(cos²x+1)^0.5dx
求不定积分!∫cos^(-1)x dx ∫cos√x dx
∫cos((x/3)-1)dx=?
∫cos(3x-1)dx
求不定积分1)∫3^(-x)*(2*3^x-3*2^x)dx 2)∫cos^2(x/2)dx
求下列不定积分.求详解(1)∫[(cos 2x)/(cos x﹢sin x)]dx; (2) ∫cot²xdx; (3) ∫{(1+2x²)/[x²(1+x²)]}dx; (4) ∫sin²(x/2)dx; (5) ∫[(cos2x)/(sin²xcos²x)]dx; (6) ∫[e^(x-4)]dx;求解
求不定积分:∫(sin x/cos^3 x)dx
求不定积分:∫(sin x/cos^3 x)dx
求不定积分∫(cos x+4x^3)dx
求积分:∫sin^2 (x) /cos^3 (x) dx
求 ∫ x^3 • cos³x dx
求∫1/(sin^4x+cos^4x)dx,
求不定积分∫{ x^4 + [1/(3x)]-cos 2x } dx∫{ x^4 + [1/(3x)]-cos 2x } dx
求不定积分∫{1/[√(x+1)+√(x-1)]}dx= ∫(sinx/cos^4x)dx= ∫ (tanx/√cosx)dx= ∫[1/(3+cosx)]dx=
∫(2x-1)cos(x-x+1)dx求积分?
求∫cos(ln(1/(1-x)))dx