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เล่นใหม่
x->-xI=int[-pi,pi](xsin(x)(pi/2-arctan(e^x))/(1+cos^2(x)))dx2I=int[-pi,pi](pi/2•xsin(x)/(1+cos^2(x)))dxI=pi/4•int[-pi,pi](xsin(x)/(1+cos^2(x)))dxthe remaining integrand is evenI=pi/2•int[0,pi](xsin(x)/(1+cos^2(x)))dxx->pi-xI=pi/2•int[0,pi]((pi-x)sin(x)/(1+cos^2(x)))dx2I=pi^2/2•int[0,pi](sin(x)/(1+cos^2(x)))dxt=cos(x)dt=-sin(x)dxI=pi^2/4•int[-1,1](1/(1+t^2))dtI=pi^2/4•(arctan(t))|[-1,1]I=pi^2/4•(pi/4-(-pi/4))I=pi^3/8
an interesting integral
x->-x
I=int[-pi,pi](xsin(x)(pi/2-arctan(e^x))/(1+cos^2(x)))dx
2I=int[-pi,pi](pi/2•xsin(x)/(1+cos^2(x)))dx
I=pi/4•int[-pi,pi](xsin(x)/(1+cos^2(x)))dx
the remaining integrand is even
I=pi/2•int[0,pi](xsin(x)/(1+cos^2(x)))dx
x->pi-x
I=pi/2•int[0,pi]((pi-x)sin(x)/(1+cos^2(x)))dx
2I=pi^2/2•int[0,pi](sin(x)/(1+cos^2(x)))dx
t=cos(x)
dt=-sin(x)dx
I=pi^2/4•int[-1,1](1/(1+t^2))dt
I=pi^2/4•(arctan(t))|[-1,1]
I=pi^2/4•(pi/4-(-pi/4))
I=pi^3/8
an interesting integral