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∫(1+sin2x)^3Cos2xDx, 区间是 {0,π/4},求定积分...

∫[π/4→π/3] 1/(sin²xcos²x) dx =4∫[π/4→π/3] 1/sin²2x dx =2∫[π/4→π/3] csc²2x d(2x) =-2cot(2x) |[π/4→π/3] =2/√3 希望可以帮到你,不明白可以追问,如果解决了问题,请点下面的"选为满意回答"按钮。

原式=0.5∫x^2 (1-cos2x)dx =0.5∫x^2dx-0.5∫x^2cos2xdx =0.5x^3/3-0.5[ x^2 *0.5sin2x-∫xsin2xdx] =0.5x^3/3-0.25x^2sin2x+0.5[-0.5xcos2x+∫0.5cos2xdx] =[0.5x^3/3-0.25x^2sin2x-0.25xcos2x+0.125sin2x] =[0.5π^3/3-0.25π]-[0] =π^3/6-π/4

3cos2x+4sin2x = 5[ (3/5)cos2x + (4/5)sin2x] =5cos(2x- arccos(3/5)) ∫dx/(3cos2x+4sin2x) =(1/5)∫dx/cos(2x- arccos(3/5)) =(1/5)∫sec(2x- arccos(3/5)) dx =(1/10)ln|sec(2x- arccos(3/5)) + tan(2x- arccos(3/5))| + C

楼主少了括号哦,老师这样写是会被扣分滴哦~等我一下我上图,求不忙采纳其他人的。。。

求不定积分∫sin²xdx解:原式=∫[(1-cos2x)/2]dx=(1/2)x-(1/2)∫cos2xdx=(1/2)x-(1/4)∫cos2xd(2x)=(1/2)x-(1/4)sin2x+C关于∫sinⁿxdx有递推公式:∫sinⁿxdx=-(sinⁿֿ¹xcosx)/n+[(n-1)/n]∫sinⁿֿ²xd...

∫ cos²x dx = ∫ (1 + cos2x)/2 dx = x/2 + (1/4)sin4x + C ∫ cos³x dx = ∫ (1 - sin²x) dsinx = sinx - (1/3)sin³x + C ∫ cos⁴x dx = ∫ (cos²x)² dx = ∫ [(1 + cos2x)/2]² dx = (1/4)∫ (1 + 2cos2x + ...

∫ (x^3+2x-1) dx =(1/4)x^4+x^2 -x + C

你代的求导公式被积函数不含 x, 这里含 x , 故你运算错误!

即对1/2sin(2x+3)求微分得到 1/2cos(2x+3) d(2x+3) 即cos(2x+3)dx 同样对d-e^(-x)微分得到e^(-x)dx 两个式子都得到了验证

∫(x^3+1)(cosx)^2dx =∫(x^3+1)[(1+cos2x)/2]dx =(1/2)∫(x^3+1)dx+(1/2)∫cos2xdx+(1/2)∫x^3cos2xdx =(1/8)x^4+x/2+(1/4)sin2x+(1/4)sin2x *x^3 -(1/4)3x^2sin2xdx =(1/8)x^4+x/2+(1/4)sin2x+(1/4)sin2x*x^3 +(3/8)cos2x*x^2-(3/4)∫xcos2xdx =(1/8...

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