求不定积分∫xcos(9 x²)dx

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

∫xcos(3x)dx=xsin(3x)/3-1/3∫sin(3x)dx(应用分部积分法)=xsin(3x)/3+cos(3x)/9+C(C是积分常数)∫xln(x+1)dx=x²ln(x+

我的解答如下：换元法令x=3/2sint,t∈[-0.5π,0.5π]dx=3/2cost带入后得到∫(1-x)/[√(9-4x^2)]dx=∫(1-1.5sint)1.5costdt/3cost=∫

∫xcos(4x^2+5)dxlet4x^2+5=tdt=d(4x^2+5)=4d(x^2)=4*2xdx=8xdxsodx=[dt/8x]∫xcos(4x^2+5)dx=∫xcostdt/8x=1/

∫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)=2xsi

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(

x^2+4x+9=(x+2)^2+5letx+2=√5tanadx=√5(seca)^2da∫5/(x²+4x+9)dx=∫[5/(5(seca)^2)]√5(seca)^2da=√5a+C

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、-1/9*(1+3*x)*e^(-3*x)+C2、1/16*cos(4*x+3)+1/16*(4*x+3)*sin(4*x+3)-3/16*sin(4*x+3)+C3、x*(-1/2*cos(x)

∫cos²xdx=∫cosxdsinx=sinxcosx-∫sinxdcosx=sinxcosx+∫sin²xdx=sinxcosx+∫(1-cos²x)dx=sinxc

∫√(x²-9)/xdx=√(x²-9)-3arcsec(x/3)+C

设x=3sect,dx=3sect*tantdt,cost=3/x,t=arccos(3/|x|),tant=√[(sect)^2-1]=√(x^2/9-1)原式=∫sect*tantdt/(|tan

∫e^sinx(xcosx-sinx/cosx^2)dx=∫e^xsinx*xcosxdx-∫e^sinxsinxdx/(cosx)^2=∫xe^sinxdsinx-∫e^sinxd(1/cosx)=

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

∫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.∫(x√x+1/x^2)dx=∫x^(3/2)dx+∫x^(-2)dx=(2/5)x^(5/2)+(-1)x^(-1)+C=(2/5)x^(5/2)-x^(-1)+C2.∫xe^xdx=∫xd(e