integral help

[tandoorichicken]

heres a funky one I can't do

\int \frac{6t \,dt}{\sqrt{1-2t^2}}


here's what I've done
I changed it so it looks like this:

\int 6t(1-2t^2)^{-1/2} dt

Then I substituted u for 1-2t^2 and got du= -4t*dt
Where do I go from here?

[sridhar_n]

Take 6 as 4 * 1.5 and put (1-2t^2) = u so that 4tdt becomes du. Then you can proceed normally. The integral now contains 1.5*u^{-1/2}du.

Sridhar

[HallsofIvy]

Rather than "seeing" that 6= 4*1.5,

Another way to look at this is: taking u to be 1-2t2 so that du= -4tdt, then tdt= -(1/4)du and 6tdt/√(1-2t2) becomes
6(-1/4)du/u1/2= (-3/2)u-1/2du