00:01
Here we have problem 30 of chapter 8 .3.
00:08
And the integral we're dealing with here is 1 -9 sign of 2x, the x from pi over 2 to 3 quarters.
00:26
Now when you have a square root or something, that's just pretty much a clue that's telling you that you need to get rid of the square root.
00:36
By using some trigonometry manipulations.
00:40
And in fact, it's not that really hard we have, because we have one minus sign of 2x here.
00:46
And, well, we have one, and whenever you see the number one in the trick, you should be thinking of cosine squared x plus sine square x.
00:59
And whenever we have a double sign of a double angle, 2 times x, this is 2 times sine, x times cosine x.
01:11
So it turns out that this whole thing is actually equal to cosine x minus sine x squared.
01:22
So we can now write our integral as, well, the same boundaries, because we didn't really do any substitution here.
01:34
And square root of the square root of cosine x minus x.
01:39
So that means we have absolute value of cosine x or sine x dx here.
01:48
Just the reminder that when you take, the reason why you have an absolute value here is if you have a square root of number a squared, this is the correct version is it's absolute value of a because a could be a negative number, but the square root of any positive number is always positive.
02:12
So we need to figure out how to get rid of the absent value here.
02:17
Luckily, it's not that troubling in our problem, because if you look at the interval here, we have pi over two going from all the way to 4 through our 4 pi.
02:31
And in this interval, sign, okay, let me illustrate this a little, real quick here.
02:39
So if, let's see, this is the graph...