00:02
All right.
00:02
So this is an interesting definite integral because you sub doesn't work.
00:08
It's not quite right for an inverse tan antiderivative or, you know, a derivative giving us inverse tan because it's squared.
00:15
So we need another trick.
00:17
And one trick is just to know your trig identity super well.
00:21
Because if you remember that tan squared of x plus one equal secant square of x, then we kind of notice that down in there kind of has a similar.
00:32
Form.
00:33
So what we can do is we can let x equal tan of x.
00:40
Or actually, we're going to make it equal to change a variable.
00:44
So let's make it equal to say tan of you.
00:47
So then d x, du equals sikin squared of u, or d x is secant squared of u, d u.
00:59
So we're going to need that because we're going to substitute in, actually both the trig identity and the substitution.
01:06
All right.
01:07
Now, when we substitute in, we have to remember to change in limits.
01:10
So when x is zero, then tan has to be some angle, well, tan of u is zero.
01:20
So we know tan of zero is zero.
01:22
So basically when x equals zero, then if tan of u is going to equal zero, u has to be equal to zero as well.
01:33
So there's a zero still there.
01:35
Now we're going to do what about when x is 1, which is the upper limit? you have to think, when is tana u equal to 1? that's when sign and cosine are the same, or when u equals pi over 4.
01:49
So our integral will be from 0 to pi over 4.
01:54
Dx will substitute as secant squared u, du, and notice, notice that i can replace x by tan of u.
02:03
So tan of u squared plus 1 squared.
02:08
All right, so now we can use that trig identity on the bottom...