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Nice, thanks!
Thank you! 😃
Very elegant👏👏👏
Love it! Great suggestion from Waarschijn
Awesome solution from Waarschijn and excellent presentation!
thanks Slavino! :)
Nice! You can use this approach to easily get the general formula Int[ x^n sin(ln x) = x^(n+1)/( (n+1)^2 + 1) { (n+1) sin(ln x) - cos(ln x)}.
excellent!! thanks :)
sin(ln(x))=(x^i-x^-i)/2iI=-i/2•int(x^(2+i)-x^(2-i))dxI=x^(3+i)/2i(3+i)-x^(3-i)/2i(3-i)+C1/(3+i)=(3-i)/10I=x^3/20i•((3-i)x^i-(3+i)x^-i)+C=x^3/10•(3sin(ln(x))-cos(ln(x)))+C
Nice, thanks!
Thank you! 😃
Very elegant👏👏👏
Love it! Great suggestion from Waarschijn
Awesome solution from Waarschijn and excellent presentation!
thanks Slavino! :)
Nice! You can use this approach to easily get the general formula Int[ x^n sin(ln x) = x^(n+1)/( (n+1)^2 + 1) { (n+1) sin(ln x) - cos(ln x)}.
excellent!! thanks :)
sin(ln(x))=(x^i-x^-i)/2i
I=-i/2•int(x^(2+i)-x^(2-i))dx
I=x^(3+i)/2i(3+i)-x^(3-i)/2i(3-i)+C
1/(3+i)=(3-i)/10
I=x^3/20i•((3-i)x^i-(3+i)x^-i)+C
=x^3/10•(3sin(ln(x))-cos(ln(x)))+C