 Let's look at the solution to question seven from our midterm exam in Calculus II Math 1220. You're asked to evaluate the integral cosine to the fourth of 2x dx. So what might we do to try to calculate the antiderivative here? Well, because we have this cosine to the fourth, it's gonna be useful to use the half angle identity here. That is cosine squared of x is equal to one half one plus cosine of 2x. If we apply that in this setting, our integral can be transformed into one half. Actually, I'll have myself there. We're gonna integrate. We're gonna, because after all, we have here a cosine squared squared. So if you use the half angle identity once, we're gonna have one half cosine, sorry, one half, one plus cosine of 4x. So it's gonna enlarge their squared dx. And so we're gonna have to do that again, right? In which case, foil this thing out, would you? You're gonna get one fourth. You're gonna get one plus two cosine of 4x plus cosine squared of 4x. In which case, we're gonna have to use the half angle identity again on this cosine square that's still there, the remnant there. And so if we do that, we're gonna end up with one fourth in front still, integral of one plus two cosine 4x, nothing changed there. Now we're gonna get a one half plus one half cosine of now 8x, like so. And so you can combine some like terms. Submittedly you're gonna get a one plus a half right here. That would give you three halves that we're gonna integrate. And then if we do that, you're gonna integrate this thing. That would then give you the three halves will integrate to become three halves x. When you distribute the one fourth onto it, you should end up with a three eighths, three eighths times x as the first part. All right, next let's consider the two cosine of 4x. The antiderivative of cosine is sine, but because of the 4x here by the U substitution, you actually have to divide by four. So you can take two divided by four, which is one half, but you have to also distribute this one fourth right here. Maybe I'm just going a little bit too fast here. That's okay, we'll back up a little bit. So we're gonna get one fourth, the integral of three halves x. Next, we're gonna get two over four sine of 4x. And then for the last one, you're gonna get one half, one half times one eighth sine of 8x. Don't forget your constant there. And so now as you distribute this one fourth onto all the pieces, one fourth times three halves x gives you three eighths x. Next, the one half here, or the two fourths actually becomes a one half. So we're gonna get one eighth sine of 4x, like so. And then lastly, we distribute the one fourth onto this one over here. We already have a one half times one eighth, which is one 16th times that by another one fourth, it's gonna give you a one 64th sine of 8x. Maybe we should have made that line a little bit longer. And then finally, don't forget your plus a constant. As we are calculating an indefinite integral, the plus c is part of it, because we're looking for the most general anti-derivative and missing the plus c is actually a forfeiture of a point there. So you wanna watch out for that. And so this would give us the anti-derivative of this function using the identity, that this half angle identity combination with some foil and other algebraic techniques.