 Let us look at a final example of an integral-evolving secants and tangents. But unlike previous examples, this one doesn't have any tangents whatsoever. We just have a secant. Now, how does one make this thing work? We don't have any tangents. So we can't really just switch things over to tangent. I mean, if we had an even number of secants, we could do that. We could take u to be tangent, and we could take du to be secant squared. And if we had an even number of secants, then we borrow two for the u substitution, and all the others can be transitioned into tangents. That's gonna be great. If you have an odd number of secants and you don't have any tangents to borrow from, you're gonna have to try a reduction technique. And so we're gonna show you how secant cube works here. Now, some things we need to remember throughout this process. So we're gonna use the fact that secant squared x is equal to tangent squared plus one. When finding antiderivatives of tangent and secant, it's best if we have both tangents and secants. So this Pythagorean identity can be useful to help us reduce down the power. So we're gonna take two of the secants and replace it not as a du, but as a tangent squared plus one. So we're gonna have one secant that's left behind. And then we're gonna transition the other ones over. Tangents squared, tangent squared x plus one, dx right there. Distribute the secant squared like so, that's gonna give us a secant x, tangent squared x plus the integral of secant x dx. And so if you look at the second one first, the secant x dx, right there we see the reduction. We reduce from the power of three to the power of one. And so one could use this technique. If you can do the first power, you can do the third power. If you can do the third power, you can do the fifth power. If you can do the fifth power, you can do the seventh power. And I'm skipping the even ones because even ones are much easier for secant here. If we have an odd power of secant, we can reduce it down like this. And we do know the antiderivative of secant. We saw that actually in a previous lecture video. That was the natural log of the absolute value of secant x plus tangent x. So we can use that fact right here. But what about this puppy, the integral of secant x tangent squared? That one takes a little bit more of a challenge, right? But it's certainly not at all approachable because we have a secant tangent. We could try to do some type of u substitution because we have a secant and a tangent. That could be a du for us, that could be a du. But the problem with that is that the u would be secant squared, which we won't have any secants left, we'll actually have a tangent left. And switching that over to a secant since there's just a single tangent would be very problematic. So it turns out that u substitution doesn't work very well for secant x, tangent x, tangent squared. What we have to do instead is employ integration by parts. Because the secant x dx is a good choice for du. So if you thought that before, that was a good choice. But the thing is, we don't want it, actually not a du. We want that to be a dv, that's classy right there. Let me fix my typo. So we want dv to be secant x tangent x dx. That way it's anti-derivative is fairly simple, it's a secant x. But then I would force that u should be tangent x, for which then du would equal secant squared x dx. So using integration by parts here, let's see what we get. You're going to get u times v, which is a secant x tangent x. Then you're going to get minus the integral. You're going to end up with a secant cubed x dx. And then we have also this natural log floating around. The natural log of secant x plus tangent x plus a constant. We don't need a plus a constant right now because the integral right here is still incorporating that plus a constant. It's kind of redundant here. So hey, this integral of secant cubed, I feel like this is data. I feel like I've seen it somewhere before. Where have we seen this before? It's right here. The integral secant cubed is what we're trying to compute right here. So remember, this is an equation, right? This expression right here is just the right-hand side of this expression right here. Secant cubed is equal to, sorry, the integral of secant cubed is equal to secant x tangent x minus the integral of secant cubed plus the natural log of something, right? If we treat this as an equation and we were to add to both sides secant cubed x dx, if we add that to both sides, on the left-hand side, we're going to have two times the integral of secant cubed x dx. And on the right-hand side, we'd have this secant x tangent x. We would have this natural log of the absolute value of secant x plus tangent x plus a constant. We do need that now, since there's no integrals that incorporate it. And so to solve for the antiderivative, we're just going to divide both sides by 2. So the integral of secant cubed x dx, this would equal 1 half the product of secant x tangent x plus 1 half the natural log of the absolute value of a secant x plus a tangent x plus an arbitrary constant. And so we started off doing integration by parts, and it seemed to turn into some type of integration by cycles variant, which we're perfectly happy with integration by cycles. It works out really nicely. And so you can see that taking the antiderivative of secant cubed can be quite challenging. The reduction's a little bit more complicated, but this is what happens when you do these reduction formulas for odd powers of secant. Do integration by parts, it's going to be some type of integration by cycle situation, and then use the reduction to help you out here. And so this technique could be mimicked to find the antiderivative of secant to the fifth, secant to the seventh, or any odd power of secant, assuming there's no tangents. If there is a tangent present, then please set du equal to be secant tangent and do a u substitution. Integration by parts, or in this case, integration by cycles, should be used as a last resort if there's no successful u substitutions you can use with these trigonometric integrals.