$\int \sin^3 x \cos x dx = $ A: $\frac{\sin^4 x}{4} +C$ B: ${\sin^4 x} +C$ C: $\frac{\cos^4 x}{4} +C$ D: $\frac{\cos^4 x}{4} +C$

求微分方程[img=364x55]17da65386dfd612.png[/img]的通解； （ ） A: - cos(2*x)*exp(x)*(x/4 - sin(4*x)/16) + C23*cos(2*x)*exp(x) + C24*sin(2*x)*exp(x) B: (3*sin(2*x)*exp(x))/32 - (sin(6*x)*exp(x))/32 - cos(2*x)*exp(x)*(x/4 - sin(4*x)/16) + C23*cos(2*x)*exp(x) + C24*sin(2*x)*exp(x) C: - sin(4*x)/16) + C23*cos(2*x)*exp(x) + C24*sin(2*x)*exp(x) D: (sin(6*x)*exp(x))/32 - cos(2*x)*exp(x)*(x/4 - sin(4*x)/16) + C23*cos(2*x)*exp(x) + C24*sin(2*x)*exp(x)

下列各组选项的两个语句运行结果不同的是? Dt[Sin[x],{x,4}] ，D[Sin[x],{x,4}]|D[Sin[x]Sin[y],x], Dt[Sin[x]Sin[y],x,Constants→y]|Dt[Sin[x]Sin[y],x] ，D[Sin[x]Sin[y],x,NonConstants→y]|Dt[x^2y^3,x,y]，D[x^2y^3,x,y]

求微分方程[img=143x21]17da5f14490e50e.png[/img]的通解,实验命令为(). A: dsolve(D2y-2*Dy+5*y=sin(2*x),x)ans =exp(x)*sin(2*x)*C2+exp(x)*cos(2*x)*C1+1/17*sin(2*x)+4/17*cos(2*x) B: dsolve('D2y-2*Dy+5*y=sin(2*x)','x')ans =cos(2*x)*(sin(4*x)/17 - cos(4*x)/68 + 1/4) - sin(2*x)*(cos(4*x)/17 + sin(4*x)/68) + C1*cos(2*x)*exp(x) - C2*sin(2*x)*exp(x) C: dsolve(D2y-2*Dy+5*y=sin(2*x),'x','y')ans =exp(x)*sin(2*x)*C2+exp(x)*cos(2*x)*C1+1/17*sin(2*x)+4/17*cos(2*x)

‍下列各组选项的两个语句运行结果不同的是‎ A: Dt[Sin[x],{x,4}] ，D[Sin[x],{x,4}] B: Dt[x^2y^3,x,y]，D[x^2y^3,x,y] C: Dt[Sin[x]Sin[y],x] ，D[Sin[x]Sin[y],x,NonConstants→y] D: D[Sin[x]Sin[y],x], Dt[Sin[x]Sin[y],x,Constants→y]

$\int {{{\sin 2x} \over {1 + {{\sin }^4}x}}} {\rm{d}}x = $ A: $\arctan (\sin x) + C$ B: $\arctan ({\sin ^2}x) + C$ C: ${\arctan ^2}(\sin x) + C$ D: $ - {\arctan ^2}(\sin x) + C$

求不定积分[img=132x48]17da6537fc8dad6.png[/img]； （ ） A: -(4*(cos(x/2)/2 + 2*sin(x/2)))/(17*exp(2*x)) B: (4*(sin(x/2)/2 + 2*sin(x/2)))/(17*exp(2*x)) C: (4*(cos(x/2)/2 + 2*sin(x/2)))/(17*exp(2*x)) D: (4*(cos(x/2)/2 + 2*cos(x/2)))/(17*exp(2*x))

求以下定积分可以使用的命令有（）。[img=199x87]1802f8c8a02c037.jpg[/img] A: x=pi/4:0.0001:5/4*pi; y=1+sin(x).*sin(x); trapz(x,y) B: f=@(x) 1+sin(x).*sin(x); q=integral(f,pi/4,5/4*pi) C: f=@(x) 1+sin(x).*sin(x); q=integral(@f,pi/4,5/4*pi) D: syms x f=1+sin(x)*sin(x); s=int(f,pi/4,5/4*pi); eval(s)

y=arcsin(4x+1)的反函数为 A: y=(sinx-1)/4, x∈R B: y=sin[(x-1)/4], x∈R C: y=sin[(x-1)/4], x∈[-π/2,π/2] D: y=(sinx-1)/4, x∈[-π/2,π/2]

$\int {{1 \over {3 + 5\cos x}}} dx = \left( {} \right)$ A: ${1 \over 4}\ln \left| {{{2\cos x + \sin x} \over {2\cos x - \sin x}}} \right| + C$ B: ${1 \over 4}\ln \left| {{{2\cos {x \over 2} + \sin {x \over 2}} \over {2\cos {x \over 2} - \sin {x \over 2}}}} \right| + C$ C: $\ln \left| {{{\cos {x \over 2} + \sin {x \over 2}} \over {\cos {x \over 2} - \sin {x \over 2}}}} \right| + C$ D: $\ln \left| {{{\cos x + \sin x} \over {\cos x - \sin x}}} \right| + C$