xcos2x不定积分视频讲解

都侧眉3198不定积分 :∫ xcos^2 xdx -
阳哪德17373902958 ______[答案] cos^2 x=(cos2x+1)/2 ∫ xcos^2 xdx =∫ x(cos2x+1)dx/2+C=(∫xcos2xdx+∫xdx)/2+C=(∫xdsin2x+x^2)/4+C =(xsin2x-∫sin2xdx+x^2)/4+C=(2xsin2x+cos2x+2x^2)/8+C

都侧眉3198求x(cosx)方dx的积分 -
阳哪德17373902958 ______[答案] ∫xcos^2 x dx =∫x(cos2x+1)/2 dx =1/2*∫xcos2xdx+1/2*∫xdx =1/4∫xcos2xd2x+1/4∫dx^2 =1/4∫xdsin2x +x^2/4 =1/4 *xsin2x-1/4∫sin2xdx +x^2/4 =xsin2x/4+x^2/4-1/8∫sin2xd2x =xsin2x/4+x^2/4+1/8∫dcos2x =xsin2x/4+x^2/4+cos2x/8 + C

都侧眉3198求方程y''+y=xcos2x的通解 -
阳哪德17373902958 ______[答案] 解∵齐次方程y''+y=0的特征方程是r²+1=0,则r=±i (i是虚数)∴此齐次方程的通解是y=C1*cosx+C2*sinx (C1,c2是积分常数)令原方程的解为y=(Ax+B)cos(2x)+(Cx+D)sin(2x)∵y'=(2Cx+A+2D)cos(2x)+(-2Ax-2B+C)sin(2x)...

都侧眉3198∫*sin2*d*求不定积分 -
阳哪德17373902958 ______[答案] ∫xsin2xdx =(-1/2)∫xdcos2x =(-1/2)(xcos2x-∫cos2xdx) =(-1/2)(xcos2x-(1/2)sin2x)+C =(1/4)sin2x-(1/2)xcos2x+C

都侧眉3198求解不定积分∫ xsin3xcosx dx -
阳哪德17373902958 ______[答案] 提供思路:先将sin3xcosx积化和差,分成两个积分,然后用分部积分法,计算. 结果:(1/2)(-(1/2)xcos2x+(1/4)sin2x)+ (1/2)(-(1/4)xcos4x+(1/16)sin4x)+C

都侧眉3198(x - sinx)^2*sinx的不定积分 -
阳哪德17373902958 ______[答案] ∫(x-sinx)^2*sinxdx=∫(x^2-2x*sinx+sin^2x)*sinxdx =-∫x²dcosx-2∫ xsinx dx+∫(1-cos^2x)sinxdx =-x²cosx+2xsinx+cosx+cos^3x/3+1/2(sin2x-2xcos2x)+c

都侧眉3198不定积分过程、、、、、、∫e^xsin^2xdx=1/2e^x - 1/5e^xsin2x - 1/10e^xcos2x+c,谢谢了、、、、、 -
阳哪德17373902958 ______[答案] ∫e^xsin^2xdx = ∫e^x(1-cos(2x))/2dx = 1/2 e^x - 1/2 ∫e^x cos(2x)dx 而 ∫e^x cos(2x)dx = 1/2 ∫e^x d[sin(2x)] = 1/2 [ e^x [sin(2x)] - ∫e^x [sin(2x)] dx ] = 1/2 [ e^x [sin(2x)] + 1/2 [ ∫e^x d[cos(2x)] ] = 1/2 [ e^x [sin(2x)] + 1/2 [ e^x [cos(2x)] - ∫e^x [cos(2x)]dx ] = 1/2 e^x ...

都侧眉3198xcos^2x的不定积分 -
阳哪德17373902958 ______ 本题是不定积分基本练习题,具体步骤如下:∫ⅹcos^2xdx=∫x(1+cos2x)dx/2=(1/2)∫xdx+(1/2)∫xcos2xdx=(1/2)*(1/2)*x^2+(1/4)∫ⅹcos2xd2x=(1/2)*(1/2)*x^2+(1/4)∫ⅹdsin2x=(1/4)ⅹ^2+(1/4)ⅹsin2x-(1/4)∫sin2xdx=(1/4)ⅹ^2+(1/4)ⅹsin2x-(1/8)∫sin2xd2x=(1/4)ⅹ^2+(1/4)ⅹsin2x+(1/8)cos2x十C.本题主要用到分部积分法和三角函数的求导公式.

都侧眉3198如何求cos3xcos2xdx的不定积分书上讲解的是把他化解为了(cos5x+cosx)/2后求的积分.我不明白怎么把cos3xcos2x化成的(cos5x+cosx)/2. -
阳哪德17373902958 ______[答案] COSx*COSy=[COS(x+y)+COS(x-y)]/2

都侧眉3198∫  xcos2xdx. -
阳哪德17373902958 ______[答案] 解∫xcos2xdx= 1 2∫x(1+cos2x)dx= 1 4x2+ 1 4∫xdsin2x = 1 4x2+ 1 4xsin2x− 1 4∫sin2xdx = 1 4x2+ 1 4xsin2x+ 1 8cos2x+c.