 Let's take a look at some more examples of trigonometric integrals involving sine and cosine. But unlike our previous examples, this time the powers of sine and cosine are going to be strictly even. We're not going to have any odd powers whatsoever. You can see in this one that we have a sine squared. Using the Pythagorean identity, we could switch that to a one minus cosine squared, but that doesn't really put us in a situation better than having a sine squared. So cosine squared is not more preferable than sine squared. And we really can't borrow a sine to do a u-substitution because that only leaves one sine left. And to transition using the Pythagorean identities, you have to have an even amount. So what does one do when you have an even amount of sines and cosines? Because sines, we have two of them. And cosines, you don't see any cosines, and that's because their power is zero there. It's an even number. So instead of the Pythagorean identity, when you have only even powers, what you wanna instead use are the half angle identities for sine and cosine. So in particular, sine squared equals one half times one minus cosine of two x, and cosine squared equals one half, one plus cosine of two x. The reason why this is useful here is if you have only even powers, this sine squared here or the cosine squared here, you have an exponent. This will translate to some linear combination using cosine of two x. Cosine of two x is much more preferable to integrate than sine squared or cosine squared of x, because you no longer have the x spawn, you just actually have a period change, which in terms of use substitutions, pretty simple to take care of. So we're gonna replace the sine squared, the bounds on the definite integral will be unfaced here. The sine squared, we're gonna switch to become a one half, one minus cosine of two x, dx right here. And the idea here is that if you have to integrate something like cosine of two x, dx, well, you just do a very basic u substitution, u equals two x, du equals two dx. Oh no, you don't have the one half. Oh, I had to put it in the wrong spot there. We need a two, and so the one half goes in front there. And so then when you integrate that, you're gonna end up with a one half. The anti-drift of cosine is sine sine of two x plus a constant. And so if your period change is off there and just divide by that constant that you need here, and so cosine of two x is fairly simple to integrate. Integrating this, we get one half, the anti-derivative of one becomes an x, the anti-derivative of negative cosine of two x, like we saw a moment ago, will be negative one half, sine of two x, and then we evaluate this from zero to pi. Now the good news is when you plug in the zero, both the x and the sine will go to zero, so they'll just disappear. That's not exactly true when you plug in a pi. Now, if you plug in pi, if we just put all the things in here, we're gonna get one half. So we get pi minus one half, sine of two pi, and then minus zero minus one half, sine of zero. Like I was mentioning a moment ago, sine of zero is equal to zero, of course zero is equal to zero. Now sine of two pi is also equal to zero, so most of this stuff is just gonna cancel out. And then you take the one half times the pi, we see the area under this curve is gonna be pi over two. So just wanna illustrate this example that if you only have an even number of sines or cosines, that's okay, use half angle identities. If one of the powers is odd, you can use the Pythagorean identity. Let's take another example, look at another example of this. This time we have sine to the fourth. How do we deal with such a thing like this? Well, we have to take it step by step. If you have only even powers, that means you can factor the exponents by taking out a two, in which case you get sine squared, squared dx. And using the half angle identity we saw on the previous slide, sine squared, you can make a substitution there, sine squared is one half, one minus cosine of two x, and this is gonna be squared here. With those half angle identities we saw on this slide, they do look very similar, right? Sine squared equals one half, one minus cosine two x, cosine squared equals one half, one plus cosine two x. Both the sine squared and the cosine squared will get a one half, one plus or minus cosine two x. It's important to keep track of where does the minus sign go? Well, the thing that you'll notice is that cosine squared gets the plus, and sine squared gets the minus. It's like cosine likes himself and hates sine. So sine's the negative one, yikes, pessimist, cosine's in love with himself, sort of a narcissist type of tendency cosine has at times. So sine squared becomes one half, one minus cosine of two x. And do be careful with your pinmanship here. When you write cosine of two x you wanna make sure it looks like a two x and not a cosine squared x. It's a very common mistake. You can put parentheses around it to help you out there. Although death by parentheses is a concern we have to have here. If we foil this thing out, the one half squared will become a one fourth. I'm gonna put that in front of the integral. And then when you foil out the one minus cosine two x squared, you end up with one minus two cosine two x plus cosine squared of two x. Like so, let me fix that. Now I could integrate one, I can integrate cosine two x but how do you integrate cosine squared of two x? That's the issue there. It turns out we wanna use the half angle identity again but this time we have to use the cosine one. And so if we do that we'll integrate the one and the cosine of two x in just a moment. So just copy down what we had before one minus two cosine two x. And so for the cosine squared of two x, that'll become by the half angle identity one half. You get another one half there. One plus, remember cosine loves himself, cosine of two two x, that is a four x. You're gonna double the angle one more time. And so now we're in a situation where we could integrate cosine of four x very similar to cosine of two x. There are some like terms you could combine right here because there's a one and a three halves you combine together. I'm just gonna hesitate to do that in just a second. We'll just integrate it. The integral of one fourth, well the coefficient's out in front. Now we have to integrate one, I don't want an integral sign now. Integrate the one that becomes an x. Integrate the negative two cosine two x. That becomes a negative cosine, sorry, negative sine of two x, like so. And then you integrate the three halves that'll be a three halves x. And then here we have to integrate a one half cosine of four x. That's actually gonna give us a one eighth cosine of four x. Sorry, I didn't change that again. One eighth sine of four x, like so. Plus a constant. I kind of botched that last one. So let's be a little bit careful what's going on. So if you consider the anti-derivative of cosine of four x, by basic use substitution, you're saying that u equals four x, du equals four dx. I don't have the four, so put it out in front by dividing by one four. That's where the one fourth came from. And then the anti-derivative would give you one fourth. The anti-derivative of cosine is sine, sine of four x plus a constant. Why a one eighth right here? Well, that's because this one half distributes onto the two pieces. Oh, my pen's not working there. There we go. The one half distributes there, so you get one half times a fourth. And so now let's distribute this one fourth through to all these pieces. But we can also combine like terms right here. Oh, where did I get that three halves there? I'm totally sorry. That should be a one half. The one half times one is one half. Integrate that to get an x there. I think I know what I did there. I was already combining dx and the one half x together. My apologies there. So you end up with, when you add the x and the one half x together, that's where the three halves came from. Times that by one fourth, you get three eighths x. Then we're gonna get a negative one fourth sine of two x. And then lastly, we end up with a one 32nd sine of four x plus a constant there. And so I made a few mistakes along the way. I was able to pick them up. Hopefully that wasn't too confusing for our viewer here, but we get our anti-derivative just like the following. Whenever you have even powers of just sines or cosines, we're gonna have to use the half angle identities to help you with your anti-derivative there.