 So I wanna do a short recap of what we have seen so far with these trigonometric integrals. If we have an integral that only involves signs and cosines, so we have some integral sine to the A of x times cosine of B to the x, dx, if you have something like that, if the power of the sign is odd, if the power of the sign is odd, you're gonna borrow one of the signs so you do a du, and that's gonna force then the u to be cosine of x. And then you can transition all the other signs into cosines using the Pythagorean identity sine squared x is equal to one minus cosine squared x. Now, if the roles are reversed, if the power of the cosine is odd, right, in that situation, you're gonna borrow one of the cosines to do a u substitution, so you want du to be negative cosine. You might need a double negative as the sign's not there. So du's gonna be negative cosine, then you want u to be sine, and then you have to transition all of the cosines into signs using the Pythagorean identity cosine squared x equals one minus sine squared. Now, what if the power of sine and cosine are both odd? Well, then you get a pick. You can do whichever one you want. Isn't life great when you have choices? Now, the third possibility is, what if this is the one we don't like as much? What if both the powers of sine and cosine, what if they're both even? Well, in that situation, you're gonna have to use the half-angle identities. Sine squared x equals one half one minus cosine of 2x, and cosine squared of x equals one half one plus cosine 2x. So what you're gonna do is if you have only even powers of sines and cosines, you have to transition using the half-angle. You're gonna replace the sine, you'll replace the sine squared with this cosine 2x, this cosine squared with the cosine 2x, multiply, foil out that thing if you have to, and this will now give you a new expression involving cosines of 2x, and then reevaluate the situation at that moment. If you have an odd cosines, then you go back to case B right here. If you after this have an even number of cosines, then you're gonna have to repeat this process, rinse and repeat until you're completed. So what this strategy does is it tells us that if we have any product of sine and cosines, who cares what the exponents are? We can calculate its anti-derivative. Now, the higher the exponents are, the harder this process is, but using u substitutions and the appropriate trigonometric identities, that is the Pythagorean and half-angle identities, we can calculate any integral involving a product of sine x and cosine x. And that'll then end our lecture 10. We'll talk some more about trigonometric integrals in the next lecture number 11, so stay tuned for that. If you have any questions on any part of this lecture or any part of this lecture series, feel free to post those questions in the comments below. I'll be glad to answer them. Click the like button, subscribe if you wanna learn more about these calculus videos in the future, and I will see you then. Bye, everyone.