求不定积分xsin^2x
求不定积分,xsin^2x -
∫x(sinx)^2dx =(1/2)∫x(1-cos2x)dx =(1/4)x^2-(1/4)∫xdsin2x =(1/4)x^2-(1/4)(xsin2x)+(1/4)∫sin2xdx =(1/4)x^2-(1/4)(xsin2x)+(-1/8)cos2x+C
∫x(sinx)^2dx =(1/2)∫x(1-cos2x)dx =(1/4)x^2-(1/4)∫xdsin2x =(1/4)x^2-(1/4)(xsin2x)+(1/4)∫sin2xdx =(1/4)x^2-(1/4)(xsin2x)+(-1/8)cos2x+C
∫x(sinx)^2dx =(1/2)∫x(1-cos2x)dx =(1/4)x^2-(1/4)∫xdsin2x =(1/4)x^2-(1/4)(xsin2x)+(1/4)∫sin2xdx =(1/4)x^2-(1/4)(xsin2x)+(-1/8)cos2x+C
∫x(sinx)^2dx =(1/2)∫x(1-cos2x)dx =(1/4)x^2-(1/4)∫xdsin2x =(1/4)x^2-(1/4)(xsin2x)+(1/4)∫sin2xdx =(1/4)x^2-(1/4)(xsin2x)+(-1/8)cos2x+C
∫xsin^2x dx =1/2∫x(1-cos2x)dx=1/2(∫xdx -∫xcos2x dx)=1/2(1/2*x^2-1/2∫x dsin2x)=1/4(x^2-xsin2x+∫sin2x dx )=1/4(x^2-xsin2x-1/2cos2x)+C
∫ xsin2x dx = -(1/2)∫ xdcos2x =-(1/2)xcos2x + (1/2)∫ cos2x dx =-(1/2)xcos2x + (1/4) sin2x + C
求xsin(2x)的不定积分 -
∫ xsin2x dx = -(1/2)∫ xdcos2x =-(1/2)xcos2x + (1/2)∫ cos2x dx =-(1/2)xcos2x + (1/4) sin2x + C
先降幂,再求积分答案如图所示,
计算下列不定积分 -
1/4)∫xd(sin2x) =(x^2)/4-(1/4)[xsin(2x)-∫sin(2x)dx] =(x^2)/4-(1/4)[xsin(2x)-(1/2)∫sin(2x)d2x] ==(x^2)/4-(1/4)[xsin(2x)+(1/2)cos2x]=……4)∫x^2 lnxdx=(1/3)∫(3x^2)lnxdx =(1/3)∫lnxd(x^3) 是什么。
e^xsin^2x的不定积分是e^x(sin2x-2cos2x)/5+C。∫e^xsin2xdx =e^xsin2x-2∫e^xcos2xdx =e^xsin2x-2e^xcos2x-4∫e^xsin2xdx =e^x(sin2x-2cos2x)/5+C 证明如果f(x)在区间I上有原函数,即有一个函数F(x)使对任意x∈I,都有F'(x)=f(x),那么对任何常数显然也有[F等会说。
不定积分过程、、、 -
∫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)] ]= 是什么。
∴原式= (1/2)e^x - (1/2)(1/5)(e^x)(2sin2x +cos2x) + C = (1/10)(5 - 2sin2x - cos2x)(e^x) + C 根据牛顿-莱布尼茨公式,许多函数的定积分的计算就可以简便地通过求不定积分来进行。这里要注意不定积分与定积分之间的关系:定积分是一个数,而不定积分是一个表达式,..