 In this video we provide the solution to question number six for practice exam number two for math 1220 In which case we have to evaluate the trigonometric integral cosine to the fourth of 2x dx So because we have these cosines to the fourth we only have cosines no signs We really can't make a u substitution work here. We're gonna want to utilize the half angle identity here So recall that if you take cosine squared of theta, this is equal to one half one plus cosine of 2 theta And so that's what we're gonna try to use here. So we rewrite this upon doing so we would get one half Well, you have to also realize that let me step back for a second We have to realize that this cosine to the fourth is gonna be a cosine squared of 2x itself quantity squared And so when we apply the half angle identity, we're gonna have to square everything in there So we get the one half one plus cosine of 4x squared So we're gonna have to foil that out the one half squared gives us a one fourth I'm gonna put that in front of the integral there So we have this one plus cosine to the 4x squared when you foil that out you get one plus two times cosine of 4x and then we have also a cosine squared of 4x DX so that second cosine squared that we've now have cosine squared of 4x We got to apply the half angle identity to that as well. So we have to apply it twice Upon doing so We're still gonna have the one the two cosine of 4x that all stays the same If you're not using parentheses on your angle here, do make sure your four looks small you wouldn't want it to accidentally turn into like a Cosine to the fourth right notice like what's the difference there cosine to the fourth x or 4x? You have to be very careful to make sure it's a coefficient and not an exponent there But for the second piece we have one half one plus cosine of 8x Like so and before calculating the antiderivative. I'm gonna take the liberty of combining some like terms Because after all we have this constant one right here this constant one half I'm also going to take the liberty of distributing the one fourth through So we're gonna have one plus one half, which is three halves times that by one fourth That's gonna give us three eighths. It's the first one there Then the next one when you take two we distribute the one fourth on to the cosine of 4x since there's a two there That'll then become one half cosine of 4x and then on the last one you distribute this one half on there But then you spit out the one fourth as well. So you get this one eighth and that situation cosine of 8x Like so and now we're ready to take the antiderivative So when we put all of this together the three eighths when you take the antiderivative become three x over eight then for the next one we have a One half cosine of 4x as you take the antiderivative of cosine and we come sine But then by the chain rule you do have to divide by that four there So we're gonna end up with a one eighth sine of 4x and Then for the last one here same basic idea when you take the derivative of the cosine it becomes a sine But because of how the chain rule works, we're gonna have to divide by eight you already have a one eighth there So you get a coefficient of one sixty fourth sine of 8x I am putting parentheses on it now to make it very clear that these are coefficients and not exponents And then don't forget your constant at the very end that then gives us the correct antiderivative