 All right, we're on one final, and I'm doing air quotes right now, final video about trigonometric integrals. This is our last lecture video for lecture 11, but we will be using trigonometric integrals in the future. This is just the last lecture that focuses on trigonometric integrals. We've seen many examples of integrals involving sines and cosines, and we've seen many integrals involving tangents and secants. Tangents and secants are like BFFs, so we like taking integrals together. What about cotangent and cosecant? They're best friends as well, and it turns out the integrals involving cotangent and cosecant are, the techniques are identical to how one does, how one works with secant and tangent. There's just a few subtle differences, of course, like the derivative of tangent, of course we know is secant squared. The derivative of cotangent is a negative cosecant squared. So do remember, there is a negative sign there that affects the u-substitution. The derivative of cosecant is well, you're gonna get a negative cosecant x cotangent x. So there's a negative sign that pops up in the u-substitution, so pay attention to that. The other parts that you're gonna need to know that are slightly different is we need to know the antiderivative of cotangent and the antiderivative of cosecant. For the antiderivative of cotangent, it works very similar to the antiderivative of tangent. You replace cotangent with cosine x over sine x dx. And you try a u-substitution. You'll take the one on the bottom, which is sine to be u. It's du is then cosine x dx, that's fortuitous. This then looks like the integral of du over u. And so you get the antiderivative of the natural log of the absolute value of u plus a constant, replacing u instead with a sine. You get the natural log of the absolute value of sine of x plus a constant, which is exactly what we see right there. So it's very similar, it's very similar to how one does the antiderivative of tangent. The antiderivative of cosecant works out very similarly. If your one is doing cosecant, whoops, lost my mouse there. If you're doing cosecant, you have to multiply by the strategic number one here. And what we're gonna do is we're gonna multiply cosecant by cosecant x plus cotangent x, and this sits above cosecant x plus cotangent x. Multiply by the strategic number one, distribute the cosecant, what you then will see next is in the numerator, you'll get a cosecant squared x plus a cosecant x cotangent x. This sits above of still the cosecant x plus the cotangent x, there's a dx right here. And like we saw in the situation of secant, there's a nice u substitution we can utilize right now. We're gonna take u to be the denominator, which is cosecant x plus cotangent x. And so then du is gonna equal, well, taking the derivatives of these things, derivative of cosecant is negative cosecant cotangent x, which is this part right here, minus the negative sign there. And then the derivative of cotangent is a negative cosecant squared x, which is this part here minus the negative sign du. Dx, excuse me. So we just need the negative sign, so let's do a double negative compensate. So in terms of the u substitution, works out really nice, you get a negative of the integral of du over u again, that becomes a negative the natural log of the absolute value of u plus a constant, in which case you could write this as the natural log of the absolute value. Remember, u here is cosecant plus cotangent. So you get cosecant x, don't forget the negative sign, negative natural log of absolute value of cosecant plus cotangent x plus a constant. And admittedly, you could leave it right here, that's a perfectly good anti-derivative, that's not how we wrote it here, we wrote it as the natural log of cosecant x minus cotangent x plus a constant. How did the negative sign get in the middle? And what's going on there? If we were to investigate that a little bit more, this is coming from the fact that coefficient out in front of a natural log is an exponent on the inside. So negative natural log is the same thing as taking the reciprocal on the inside. So you get this cosecant x plus cotangent x. This is all under one, plus your constant. So you can rewrite it in a manner similar to this. And so then how do we get the other thing on the top? Well, this comes from using some Pythagorean identities. We're gonna do a similar trick to what we did a moment ago. We're gonna multiply the bottom by cosecant x minus cotangent x. And we have to do that to the top as well. cosecant x minus cotangent x. It's similar to what we started off with, but for different reasons. In this situation, we're trying to utilize the Pythagorean identity for cosecant and cotangent. You get the natural log of the value on the top. You're gonna get a cosecant x minus a cotangent x. That looks familiar. On the bottom, if you multiply that out, you're gonna get a cosecant times cosecant. So you get a cosecant squared. You're gonna get a cosecant times a cotangent. That's negative. You're gonna, that's when you multiply these ones together. You're gonna get a cotangent times a cosecant, which is positive, those cancel out. And then you end up with a negative cotangent squared plus our constant. And now this is where the Pythagorean identity comes into play. It is true that one plus cotangent squared is equal to cosecant squared. So therefore, if you take the difference, this is equal to one. And this is where the simplification comes from. If we simplify this, this would end up with the natural log of the absolute value of cosecant x minus cotangent x plus a constant. And so that's why we generally just take this to be the anti-derivative. It's true, there's some steps and the simplification involves some trigonometric identities. This one might be worth just memorizing our recording it in a place where you can find easy access to it. And so in the homework, you'll see some anti-derivatives that involve cosecant and cotangent. You'll handle them basically the same way that you handle secant and tangent, but you'll need some slightly different identities as we've seen in this video right here. And like I said, this brings us to the end of lecture 11. Thanks for watching everyone. Like usual, if you have any questions about any of these lectures, feel free to reach out to me. You can email me directly or you can just post comments here on the YouTube page and I'll reply to them directly right here. If you like these videos, please click the like button, subscribe to see future videos in the future. If you wanna see the script for these videos, read the description, there's a link to a written description or some lecture notes for which these videos are based upon. And I will see you next time everyone. Keep on calculating, bye.