dx (sin^6x cos^6x)的不定积分

错,前一项∫xcos(x^2)dx=1/2∫2xcos(x^2)dx=1/2∫cos(x^2)dx^2=1/2sin(x^2)+c1,根据分部积分法,后一项∫-sin(x^2)/x^2dx=∫sin(

∫(sin2x)/(sin²x)dx=∫(2sinxcosx)/(sin²x)dx=2∫cosx/sinxdx=2∫(1/sinx)d(sinx)=2ln|sinx|+C_____

∵[xcos(x+y)+sin(x+y)]dx+xcos(x+y)dy=0==>xcos(x+y)dx+xcos(x+y)dy+sin(x+y)dx=0==>xcos(x+y)(dx+dy)+sin(

f(x)=sin2ωx+√3cos2ωx=2sin(2ωx+π/3),两对称轴之间的最小值为π/2即半个周期,则周期为π=2π/2ω,所以w=1,所以f(x)=2sin(2x+π/3),f(α)=2s

求微分方程ycos(y/x)=[(x²/y)sin(y/x)+xcos(y/x)]dy/dx的通解令u=y/x,则y=ux,dy/dx=u+xdu/dx,代入原方程得：uxcosu=[(x/

合并同类项么,很简单的只要你愿意去做左边=cos*x（cos*y+sin*y）+sin*x（cos*y+sin*y)=cos*x+sin*x=1=右边

原式=∫4dx/(2sinxcosx)²=4∫dx/sin²2x=2∫csc²2xd2x=-2cot2x+C

利用半角公式如图降次计算.经济数学团队帮你解答,请及时采纳.

intln(tanx)/(sinxcosx)dx=intln(tanx)*cosx/sinx*1/cos^2xdx=intln(tanx)*1/tanxd(tanx)=intln(tanx)d[ln(

∫(cos²x-sin²x)/(sin²xcos²x)dx=∫cos2x/[(1/2)²sin²2x]dx=2∫1/sin²2xd

f(x)=a(sin²x+cos²x)(sin^4x-sin²xcos²x+cos^4x)+b(sin^4x+cos^4x)+6sin^2xcos^2x=a(s

看：(对不起,第一条的变数全部都是t,刚才做的时候不小心把t打错作x了)

∫(1/sin²xcos²x)dx=∫(sin2x+cos2x/sin²xcos²x)dx=∫(1/sin²x+1/cos²x)dx=-co

用分部积分∫xcos（x/2）dx=2∫xcos(x/2)d(x/2)=2∫xdsin(x/2)=2xsin(x/2)-2∫sin(x/2)dx=2xsin(x/2)-4∫sin(x/2)d(x/2)

原式=0.5∫cos(1+x²)d(x²)=0.5sin(1+x²)+C再问：能给下过程么？3Q再答：这都是可以直接积分的，xdx=0.5d(x²)=0.5d(

∫arctan(1/x)dx=∫(x)'arctan(1/x)dx=xarctan(1/x)-∫x*{1/[1+x^(-2)]}*[-1/x^2]dx=xarctan(1/x)+∫1/(x+1/x)d

∫xcos(x^2)dx=∫cos(x^2)(xdx)=∫cos(x^2)(d(x^2)/2)=(1/2)∫cos(x^2)d(x^2)=(1/2)sin(x^2)+C

∫xcos(x/3)dx=3∫xdsin(x/3)=3xsin(x/3)-3∫sin(x/3)dx+C=3xsin(x/3)+9cos(x/3)+CC为任意常数

1/[(sinx)^3(cosx)^3]=[sinx/(cosx)^3]+(2/sinxcosx)+[cosx/(sinx)^3]∫(1/sin³xcos³x)dx=[(1/2)/

∫sin^2xcos^3xdx=∫sin^2x(1-sin^2x)dsinx=∫sin^2x-sin^4xdx=(1/3)sin^3x-(1/5)sin^5x+C不是让你求助我吗.再问：∫sin^2x