Volume of Prisms
2 answers:
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I'm assuming you're talking about the indefinite integral
and that your question is whether the substitution
would work. Well, let's check it out:
which essentially brings us to back to where we started. (The substitution only served to remove the scale factor in the exponent.)
What if we tried
next? Then 
, giving
Next you may be tempted to try to integrate this by parts, but that will get you nowhere.
So how to deal with this integral? The answer lies in what's called the "error function" defined as
By the fundamental theorem of calculus, taking the derivative of both sides yields
and so the antiderivative would be
The takeaway here is that a new function (i.e. not some combination of simpler functions like regular exponential, logarithmic, periodic, or polynomial functions) is needed to capture the antiderivative.
and that your question is whether the substitution
which essentially brings us to back to where we started. (The substitution only served to remove the scale factor in the exponent.)
What if we tried
Next you may be tempted to try to integrate this by parts, but that will get you nowhere.
So how to deal with this integral? The answer lies in what's called the "error function" defined as
By the fundamental theorem of calculus, taking the derivative of both sides yields
and so the antiderivative would be
The takeaway here is that a new function (i.e. not some combination of simpler functions like regular exponential, logarithmic, periodic, or polynomial functions) is needed to capture the antiderivative.