 So we've seen examples on what to do with an integral involving tangents and secants if we have a If we have an even number of secants, it's pretty okay If we have an odd number of tangents and we can afford to give up one secant We're also going to be okay But I want to explore some situations where it's not as simple certainly doable Like if you look at this integral right here take the integral of tangent cubed The problem here is we have an odd number of tangents But we don't even have a single secant to offer up for our du right? We might want to say du equals secant tangent, but we don't have a secant to do it So that doesn't work and also we do have an even number of secants There's zero but we need at least two to do the u substitution. So it's u substitution doesn't really work in this situation So what can we do? So in this situation, we have a tangent cubed and no secants whatsoever We're going to try to do some type of reduction technique That is we're going to kick the can down the road from tangent cubed into something else for this to work We're going to have to remember. What is the anti derivative of tangent, which we have done before Click the video link you see right now to see the derivation of this anti derivative here But we've seen before that the anti derivative of tangent x is the natural log of the absolute value of secant x plus Plus a constant we're going to utilize that and the basic derivation that comes from the fact that tangent is sine over cosine You can try to make that substitution right here, but I'm going to I'm going to try a simpler approach So what we're going to do is we're going to break this up because we still do know I mean it is still a fact that One plus tangent equals secant squared are more importantly tangent squared Equals secant squared x minus one so what we're going to do is in order to reduce a tangent to higher powers We're going to borrow two of the tangents and switch those tangents into a secant squared minus one So doing that we're going to rewrite This we borrow one of the tangents and then like I said the other tangents squared We're going to reduce it using what we're going to substitute using the Pythagorean identity So if we do that we get a tangent x we're going to get a secant squared x minus one DX and now you're going to distribute the tangent across this difference right here on the first one You end up with a tangent x secant squared x dx and on the second one you're going to end up with a Tangent x dx like so now for the first one because you have a tangent and a secant together now You could try to do you equal secant Du equals tangent secant you also in fact could you work with du being secant squared if you prefer The point is because you have tangents and secants together There is a possibility of doing some use substitution on the first integral and in fact we've already done This integral I'm not going to do it again, but we saw previously that this first integral becomes one half tangent squared x Plus a constant We don't need that constant here right now Because we'll just stack it at the very end of this integral there it is And then we have to do this negative tangent of x right and like I said above the Anti-derivative of tangent of x is going to be the natural log of the absolute value of secant x Clearly I didn't need so much space between my plus c there, but there she blows we have it And so what we see here is this really nice trick of Reduction of tangents if you have a power of tangent Regretting putting the plus c so far out there if you have a power of tangent like so here we had like a tangent cube But this technique would also work for a tangent of the fourth and in the fifth tangent squared Tangents of the five thousand two hundred eighty power this idea is we can reduce the tangent downs We can reduce it using this identity right here Replace two of the tangents with the secant squared minus one Distribute the remaining tangents you'll then have a tangent and a secant squared for what you can do a u substitution it'll be just fine and Then you'll have a fewer tangents left behind so like again if we did something like take tangent To the 17th of x dx what you do is you replace this with tangent to the 15th Times secant squared x minus 1 dx You distribute on the first integral You would have This tangent to the 15th secant squared d secant squared x dx minus Tangents to the 15th dx so notice here what happens is for the first one you can get away with a u substitution equals tangent du Equals secant squared dx in which case that one's not too hard to do It'll just be a power function and now here you just have one You have two less tangents than you had before so reduce the power down And so this process would eventually terminate it takes several steps I'm not going to give you one with this big of a power But this demonstrates that we can always reduce down powers of tangent So even if you have no secants available we can always artificially insert Secants into the problem using the Pythagorean identity and then we reduce using some type of induction argument to finish up this anti-derivative