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求∫2xCos(x^2+1)Dx的不定积分.

凑微分即可

∫xcos2xdx =(1/2)∫xdsin2x =(1/2)x.sin2x -(1/2)∫sin2xdx =(1/2)x.sin2x +(1/4)cos2x + C

可以用降幂公式啊,∫ cos²2x dx=1/2 * (1+cos4x)dx=1/2*x+1/2*1/4*sin4x+c,楼上用的是换元法,希望能给你提供另外一种解答~

∫cos2xdx =∫cos2xd(1/2)2x =-1/2sin2x+C

这个积分很少见啊,你在哪弄的啊,我做出来不是一个具体的函数,是一串表达式,大概是 ∫cos(x^2)dx=1/2*1/x*sinx^2-1/4*x^(-3)*cosx^2-3/8*x^(-5)*sinx^2-……-(-1)^(n-3)*【1*3*5*……*(2n-3)】/2^n*x^(1-2n)*|sin([(n-1)pi/2]-x^2)| (n趋于正无穷...

你好!可以如图改写并套用基本积分公式得出答案。经济数学团队帮你解答,请及时采纳。谢谢!

答: ∫ xcos(x/2) dx =2∫ xcos(x/2) d(x/2) =2∫ x d [sin(x/2)] =2xsin(x/2)-2∫ sin(x/2) dx =2xsin(x/2) +4cos(x/2)+C

∫x^2[cos(x/2)]^2dx =(1/2)∫x^2(1+cosx)dx =(1/2)∫x^2dx+(1/2)∫x^2cosxdx =(1/6)x^3+(1/2)∫x^2d(sinx) =(1/6)x^3+(1/2)x^2sinx-(1/2)∫sinxd(x^2) =(1/6)x^3+(1/2)x^2sinx-∫xsinxdx =(1/6)x^3+(1...

显然1+cos2x=2(cosx)^2 那么 原积分 =∫1/2(cosx)^2 dx =0.5 *∫1/(cosx)^2 dx =0.5tanx +C,C为常数

∫(x-1)cos2xdx =∫(xcos2x-cos2x)dx =∫xcos2xdx-∫cos2xdx =1/2∫xdsin2x-∫cos2xdx =1/2(xsin2x-∫sin2xdx)(分部积分法)-∫cos2xdx =1/2xsin2x-1/2∫sin2xdx-∫cos2xdx =1/2xsin2x-1/4∫sin2xd2x-∫cos2xdx =1/2xsin2x+1/4cos2x-1/2sin2x 你是第三步错了...

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