# u-substitution

joe shmo
Posers and Puzzles 08 Dec '08 01:51
1. joe shmo
Strange Egg
08 Dec '08 01:511 edit
so if you were going to evaluate the indefinite integral that would need a variable factored in by the u subtitution method how is it done without expansion and term by term integration?

for instance:

Integrand: (2x^2)^2

let U = 2x^2
dU=(4x)dx

the problem is that the varaible "x" cannot be brought outside the inteegral so how does one "massage it".. so to speak

perhaps this method isnt workable for this type of integrand?
2. PBE6
Bananarama
08 Dec '08 05:44
Originally posted by joe shmo
so if you were going to evaluate the indefinite integral that would need a variable factored in by the u subtitution method how is it done without expansion and term by term integration?

for instance:

Integrand: (2x^2)^2

let U = 2x^2
dU=(4x)dx

the problem is that the varaible "x" cannot be brought outside the inteegral so how does one "massage it".. so to speak

perhaps this method isnt workable for this type of integrand?
Note quite sure what you mean...are you asking what you can do when your u-substitution produces a term that includes the original variable? Often times you can simply solve for the value of the original variable in terms of the new one by using your original substitution equation.

In your example for instance, since U = 2x^2, we have x = (U/2)^0.5 and therefore dx = 1/(8U)^0.5. Subbing this back in, we have:

INT((2x^2)^2)dx = INT(u^2)dx = INT((u^2)/(8U)^0.5)du = INT((u^1.5)/(8^0.5))du

This solution to this integral is (2/5)*(1/8^0.5)*u^(5/2) + C = (4/5)*x^5 + C, which is the same solution you get by expanding the original equation and integrating normally.