1. ## Evaluating Definite Integrals

$\displaystyle \int_{-2}^{2} \frac{3dt}{4+3t^2} \, dt$

is my u=4+3t^2?

2. ## A different method

that won't work because, then $\displaystyle du=6t dt$.

Since there is no t on the upper side of the fraction (sorry I can't remember the name that is called) it will be a dead end.

try using the identity,
$\displaystyle \frac{d\tan{^{-1}}x}{dx}=\frac{1}{1+x^2}$

3. ## Re: Evaluating Definite Integrals

Originally Posted by purplec16
[tex]int_{-2^2}{\frac{3dt}{4+3t^2}}[\tex]

is my u=4+3t^2?
Why not try $\displaystyle t=\frac{2}{\sqrt{3}}\tan\theta$

4. ## Re: A different method

that won't work because, then $\displaystyle du=6t dt$.

Since there is no t on the upper side of the fraction (sorry I can't remember the name that is called) it will be a dead end.

try using the identity,
$\displaystyle \frac{d\tan{^{-1}}x}{dx}=\frac{1}{1+x^2}$
Yes, but using t's instead of x's, of course .....

@OP: And since the integrand is even, you can integrate from 0 to 2 and then double the resulting answer.

5. ## Re: A different method

that won't work because, then $\displaystyle du=6t dt$.
$\displaystyle \frac{d\tan{^{-1}}x}{dx}=\frac{1}{1+x^2}$