00:01
In this problem, we have to compute the definite integral from 0 to 2 over square root of 3, 1 divided by 4 plus 9x square dx.
00:16
So let's recall the relevant formula.
00:18
The formula we are going to use that the antiderivative of a square plus x square is equal to 1 over a times of.
00:30
Arc 10 x divided by a.
00:34
So now we see that we need x square, but instead we have 9 x squared.
00:42
So how do we mean this situation? so we can rewrite the expression as 0 to 2 divided by square root of 3, 1 divided by, let's take the 9 outside.
00:56
This becomes 4 divided by 9 plus x squared t x squared.
01:02
So let's take the 1 over 9 out so this becomes 0 2 divided by square root of 3 1 divided by 2 third square plus x square d x now if i apply the formula over there so this becomes 1 over a this is nothing but 3 over 2 times arc 10 of x divided by a this becomes 3x divided by 2 and then the lower limit is 0 and the upper limit is 2 over square root of 3 and i see that i can cancel this 3 so this becomes 1 over 6 then i have arc 10 of 3 now the upper limit is 2 over square root of 3 minus arc tan of 0 and arc ton of 0 i know that that is equal to 0.
02:09
I can cancel the 2.
02:11
Then this becomes 1 over 6...