00:01
So for this integral, what we're going to do is we're going to put this tan squared theta.
00:05
We're going to change that into sine squared divided by cosine squared theta, knowing just that tan theta is equal to sine divided by cosine.
00:14
So this would be equal to the integral from 0 to pi over 4 of cosine squared theta times sine squared theta divided by cosine squared theta times d theta.
00:30
And we can see here that these cosine squareds are going to cancel.
00:35
And what we're left with is just the integral from 0 to pi over 4 of sine squared theta d theta.
00:44
And what we want to do with this now is we're going to use an integration by parts.
00:49
And we're going to let u equal sine squared theta.
00:53
So then du is equal to 2 sine theta, cosine theta, and dv here would just be equal to d theta, so v is equal to theta.
01:08
And so we're going to be using this substitution to figure out the value of this integral.
01:16
So we have u times v, which is going to be theta times sine squared theta, and then minus the integral of v.
01:23
D .u.
01:25
The thing that we want to do before we go ahead and make or do this integral is we want to make a substitution in for this 2 -sign theta -cosin -theta and we're going to use a trigonomic identity that tells us that sine -2 -theta is equal to 2 -sine -theta cosine -theta.
01:44
So we're going to be plugging in sine -2 -theta in for du and so we're going to have theta times sine 2 -theta times d -theta.
01:58
And then for this integral, what we're going to do is another integration by parts.
02:03
And we're going to let u equal theta and d -u equal d -theta.
02:09
So then d -v is equal to sine 2 -theta, and v here would be equal to negative cosine 2 -2 -2.
02:21
And so now we just need to make another substitution.
02:24
And so we still have this theta -times sine -squared theta term.
02:28
And then we're going to have minus in parentheses u times v...