$\displaystyle \int t^2 (1+t^2)^\frac {1}{2}dt$ How can I do this? If I use a substitution, u=1+t^2 => t = (u-1)^1/2 then du/dt = 2t Then it is more complex. What to do?
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try using u = tan x
Originally Posted by Danshader try using u = tan x Ya. I get it. Thanks
Originally Posted by geton $\displaystyle \int t^2 (1+t^2)^\frac {1}{2}dt$ How can I do this? If I use a substitution, u=1+t^2 => t = (u-1)^1/2 then du/dt = 2t Then it is more complex. What to do? This is actually a very difficult integral...Deshand was right you cant use algaebraic substitution to a happy end
Originally Posted by geton $\displaystyle \int t^2 (1+t^2)^\frac {1}{2}dt$ How can I do this? If I use a substitution, u=1+t^2 => t = (u-1)^1/2 then du/dt = 2t Then it is more complex. What to do? With trig sub you hget $\displaystyle \int\tan^2(\theta)\sec(\theta)d\theta$
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