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

[tandoorichicken]

heres a funky one I can't do
[tex]
\int \frac{6t \,dt}{\sqrt{1-2t^2}}
[/tex]

here's what I've done
I changed it so it looks like this:
[tex]
\int 6t(1-2t^2)^{-1/2} dt
[/tex]
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 [tex](1-2t^2) = u [/tex] so that [tex]4tdt[/tex] becomes [tex]du[/tex]. Then you can proceed normally. The integral now contains [tex]1.5*u^{-1/2}du[/tex].

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/u^1/2= (-3/2)u^-1/2du