求不定积分∫xsin^2xdx
求不定积分 xsin^2xdx -
∫xsin^2xdx = (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^2xdx=1/4∫2xsin^2xd2x 令t=2x =1/4∫tsin^tdt=1/4(sint-tcost)因此∫xsin^2xdx=1/4(sin2x-2xcos2x)
∫xsin^2xdx = (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^2xdx=1/4∫2xsin^2xd2x 令t=2x =1/4∫tsin^tdt=1/4(sint-tcost)因此∫xsin^2xdx=1/4(sin2x-2xcos2x)
朋友,您好!详细过程如图rt所示,希望能帮到你解决问题,
用分部积分法∫xsin^2xdx=0.5∫x(1-cos2x)dx =0.5∫xdx-0.25∫xdsin2x =0.25x^2-0.25xsin2x+0.25∫sin2xdx =0.25x^2-0.25xsin2x-0.125cos2x+C
求不定积分 要步骤 xsin^2xdx -
过程见图:经过验证,是正确的。
解:因为f(x)=xsin²x/(1+cos²x )在(1,1)上是奇函数,所以∫【1,1】xsin^2xdx/1+cos^2x=0
求不定积分∫e^xsin^2xdx -
∫ e^xsin²x dx =(1/2)∫ e^x(1-cos2x) dx =(1/2)e^x - (1/2)∫ e^xcos2x dx (1)下面计算:∫ e^xcos2x dx =∫ cos2x d(e^x)分部积分=e^xcos2x + 2∫ e^xsin2x dx =e^xcos2x + 2∫ sin2x d(e^x)再分部积分=e^xcos2x + 2e^xsin2x - 4∫ 还有呢?
先降幂,再求积分就行,答案如图所示,
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)] ]= 说完了。
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