作业帮二重积分xcos(x)
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用直线x=1将区域分成D1与D2两部分,然后分别积分即可（如图）最后计算需要用分部积分法求出原函数,然后用微积分基本定理即牛顿-莱布尼茨公式求解
y'=cos(1/x)+x(-sin(1/x))(-1/x^2)=cos(1/x)+1/x*sin(1/x)
∫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+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(
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.
f'(x)=x'cos3x+x*(cos3x)'=cos3x+x(-sin3x)*(3x)'=cos3x-3xsin3x
(π:0)表示上下限I=∫(π:0)xdx∫(x:0)cos(x+y)dy=∫(π:0)[(sin(x+y)|(x:0)]xdx=∫(π:0)(xsin2x-xsinx)dx=-x(cos2x)/2+
=1/(ln10)[e^(-x)*cos(pix)]*[e^(-x)cos(pix)]'=1/(ln10)[e^(-x)*cos(pix)]*[-e^(-x)cos(pix)-e^(-x)(-pisi
∫(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(
∫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
被积函数是开口向下的椭圆抛物面,它与xoy面的交线是椭圆:4x^2+y^2=4 即 x^2+y^2/2^2=1. 如上图.易知 z=4-4x^2-y^2,当&nbs
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