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badtwistoffate Sep26-05 10:41 PM

does this make sense
in int[6/(x^2+3)^2] i use 3tanU=x, dx=3sec(u)^2 du
using that i get to .... int[1+cos2u]= u + sin2u/2 = u +sin u cos u... but subing back in with trig doesnt give me the right answer... any help with this method?

Benny Sep26-05 11:01 PM

[tex]I = \int {\frac{6}{{\left( {x^2 + 3} \right)^2 }}dx} [/tex]

Instead of setting x = 3tan(u) try...

x = \sqrt 3 \tan \theta \Rightarrow dx = \sqrt 3 \sec ^2 \theta d\theta

badtwistoffate Sep26-05 11:03 PM

darn it, after looking at my work i was afraid someone would say that... as i didnt relize that would be better until i finished it and noticed it was coming out right.... does that mean you cant do it my way:-/

Benny Sep26-05 11:49 PM

There's really no reason why it can't be done your way. It's just that some substitutions will require you to evaluate much more complicated integrals than if you were to use a substitution which obviously simplifies the integrand.

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