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

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

你写的式子感觉都不对,有歧义 1.猜测你想表达的意思是: ∫ x cos(2 x) dx = 1/2 x sin(2 x)-1/2 ∫ sin(2 x) dx 令 u = 2 x 则 du = 2 dx: = 1/2 x sin(2 x)-1/4 ∫ sin(u) du = (cos(u))/4+1/2 x sin(2 x)+C 代回 u = 2 x: = 1/4 cos(2 x)+x sin...

∫(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)x^4+x/2...

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...

∫3^x*5^(-x)dx =∫(3/5)^xdx =(3/5)^x/ln(3/5)+C ∫1/((sin^2x)*(cos^2x))dx =∫4csc^2(2x)dx =-2cot(2x)+C

利用公式降幂。 ∫sin²x dx=∫(1-cos2x)/2 dx=1/2x-sin2x/4+C。 ∫cos²x dx=∫(1+cos2x)/2 dx=1/2x+sin2x/4+C.

积化和差∫sinxsin2xsin3xdx=1/2∫(cosx-cos3x)sin3xdx=1/2∫cosxsin3xdx-1/2∫cos3xsin3xdx=1/4∫(sin2x+sin4x)dx-1/4∫sin6xdx=-1/8cos2x-1/16cos4x+1/24cos6x+C数学软件验算:

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...

1)∫√(2+3x)dx t=2+3x,x=1/3*t-2/3,dx=1/3dt )∫√(2+3x)dx=St^(1/2)*1/3dt=1/3*2/3*t^(3/2)+c=2/9*(2+3x)^(3/2)+c 2)∫4/(1-2x)^2dx t=1-2x,x=-1/2*t+1/2,dx=-dt )∫4/(1-2x)^2dx=S4/t^2 *(-dt)=-4St^(-2)*dt=4/t+c=4/(1-2x)+c 3)∫sin3xdx t=3x...

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