Question

∫_2^(-2) {(3dt/(4+3t^2))}

Answer 1

http://integrals.wolfram.com/index.jsp

Answer 2

\[\int\limits_{-2}^{2}{\frac{3dt}{4+3t^2}}\]

Answer 3

i dont want to know the answer i need to know how to do it

Answer 4

lol

Answer 5

here we go:
\[1/4 * 3/ (4/4 +(\sqrt{3}x/2)^{2}=3/4 * 1/(1+(\sqrt{3}x/2)^{2}\]

Answer 6

this is under integral

Answer 7

I'm feeling an arctan flavor here:\[\frac{3dt}{4+3t^2}=\frac{3dt}{(4)(1+\frac{3t^2}{4})}=\frac{3}{4}\cdot\frac{dt}{1+(\frac{\sqrt{3}t}{2})^2}\]Move the 3/4 to the outside of the integral, "balance" the squared portion, use arctan...

Answer 8

now you have:
\[3/2\int\limits_{ }^{}d(\sqrt{3}x/2)/(1+(\sqrt{3}x/2)^{2} *2/\sqrt{3}=\sqrt{3}/2 \tan ^{-1}(\sqrt{3}x/2)\]

Answer 9

put your (-2,2) & calculate

Answer 10

are you OK with this?
generally it's substitution u=sq.rt(3) x/2

Answer 11

its not really clear what u did but i'll write it out thank you inik and londovir

Answer 12

The main thing is if you ever see an integral with no variable in the numerator (perhaps just a constant number), but with a squared variable in the denominator, it's often inverse trig...when it's a + in the denominator, think inverse tangent (or arctan)...

Answer 13

or we didnt learn this yet...okay thank you