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?
[tex]I = \int {\frac{6}{{\left( {x^2 + 3} \right)^2 }}dx} [/tex] Instead of setting x = 3tan(u) try... [tex] x = \sqrt 3 \tan \theta \Rightarrow dx = \sqrt 3 \sec ^2 \theta d\theta [/tex]
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:-/
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.