 Let's take another look at an example of an integral involving both tangent and secant here We saw in a previous example that if your power of secant is an even number You can borrow two of those secants to be your du. You said u to be tangent of x and you can do a u substitution What I want to do is show you what you can do if your Power of secant is in fact odd if your power is odd The idea is you don't want to use this one secant squared what you want to do And the reason for that is if you have an even amount you take away To you'll be left with an even amount and you can then convert secants into tangents using the Pythagorean identity If you have an odd amount and you borrow to you'll have an odd amount And then you can't do the Pythagorean identity to even there on the other hand if you use if you set u to be tangent secant I should say du to be tangent secant. We take du to be a tangent x secant x dx Then in that situation you should set u equal to be u to be Spacing out here secant of x what would happen in the previous situation So let's kind of break it up into smaller pieces the tangent to the fifth secant to the seventh Well, what we're going to do is we're going to borrow one of this tangents and one of the secants So we get tangent to the fourth theta We're going to get secant to the sixth theta and then we have this tangent theta secant theta d theta Right here And so notice this tangent secant d theta. We're setting this aside to be our du I noticed I was writing x earlier with our variables. That was mistake on my part. They should be theta's So the tangent theta secant theta d theta will become our d theta What do you do with the rest of it? Well secant to the sixth should be fairly clear secant to the sixth should just be u to the sixth What about the tangent to the fourth in that situation? Well utilizing the Pythagorean identity above here notice that tangent squared I should say in this case tangent squared theta is equal to secant squared theta minus one Making that substitution tangent to the fourth is going to become all tangent to the fourth is just tangent squared squared So we're going to get a secant squared theta minus one. That's a tangent squared And then if we square that we'll get a tangent to the fourth And so recognizing, of course every secant becomes a u This will become a u squared right there And so employing this u substitution Our integral can be changed into u squared minus one squared We get u to the sixth du And so this is a nice polynomial expression We can multiply this thing out first foil out the u squared minus one squared If we do that we end up with a u to the fourth minus two u squared plus one u to the sixth du I then would recommend distributing the u to the sixth onto all three pieces Doing that we end up with the integral of u to the tenth Minus two u to the eighth Plus u to the sixth now. Don't be overwhelmed if the power seems so big. Oh, no u to the tenth. Oh dear Well for the power rule makes really no difference I mean if our power is one million then we integrate the power will become million and one So just just employ the power rule as you've seen with the antiderivatives in the past We would end up with one 11th u to the 11th minus two ninths u to the ninth plus one seventh u to the seventh plus a constant and now replace the u with secant of theta Which was our original substitution there in which case we get one 11th Not u secant to the 11th theta Then we get minus two ninths secant to the ninth theta plus one seventh one seventh secant to the seventh theta plus a constant and here's our antiderivative So we've now seen in this example plus the previous one if the power of secant is even what to do In this one I have to add a little bit of a caveat, right? We I said this is what we do when secant has power is odd, but it also only works if tangent's power is also odd It can get a little bit funky otherwise But if you have an odd number of tangents and an odd number of secants this this will work out very well for us