Solving an Intricate Integral: \int 6t(1-2t^2)^{-1/2} dt

[tandoorichicken]

heres a funky one I can't do
<br /> \int \frac{6t \,dt}{\sqrt{1-2t^2}}<br />

here's what I've done
I changed it so it looks like this:
<br /> \int 6t(1-2t^2)^{-1/2} dt<br />
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/&radic;(1-2t²) becomes
6(-1/4)du/u^1/2= (-3/2)u^-1/2du