# Integral help

1. Dec 10, 2003

### 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?

2. Dec 10, 2003

### 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

Last edited: Dec 11, 2003
3. Dec 11, 2003

### HallsofIvy

Staff Emeritus
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/u1/2= (-3/2)u-1/2du