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dx/dy=?
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Integral Substitution...
Heya people,
I was wondering if someone here could point me in the right direction, as the book I am reading on Integration isn't very thourough, and I don't really have anyone else to ask.
Basically, I am reading up on u-substitiution regarding integration, but I am not really sure of the finer points.
The texts says that I can ONLY use substituiton if the derivative of the 'u' is present in the integral-equation.
My Question:
When you have an integral to evaluate, and its complicated enough to have to use u-substituiton, but there is not any [tex] g'(x) [/tex] for your [tex] g(x) [/tex] in the integral,
can we factor in the required derivative of our 'inside' function to the equation before evaluating the integral to make things easier?
Ie) Does this mean, (for simplicity,)
If I have:
[tex]\int sin(x^4) dx[/tex]
and I want to make my [tex]\ u =x^4[/tex]
then, can I somehow factor in the 4x^3 to make this work?
Cheers.
Heya people,
I was wondering if someone here could point me in the right direction, as the book I am reading on Integration isn't very thourough, and I don't really have anyone else to ask.
Basically, I am reading up on u-substitiution regarding integration, but I am not really sure of the finer points.
The texts says that I can ONLY use substituiton if the derivative of the 'u' is present in the integral-equation.
My Question:
When you have an integral to evaluate, and its complicated enough to have to use u-substituiton, but there is not any [tex] g'(x) [/tex] for your [tex] g(x) [/tex] in the integral,
can we factor in the required derivative of our 'inside' function to the equation before evaluating the integral to make things easier?
Ie) Does this mean, (for simplicity,)
If I have:
[tex]\int sin(x^4) dx[/tex]
and I want to make my [tex]\ u =x^4[/tex]
then, can I somehow factor in the 4x^3 to make this work?
Cheers.
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