 The trigonometric functions are extremely important in pre-calculus, so, no surprise, the integrals of trigonometric functions are very important in calculus. If you only remember one thing about the trigonometric functions, you should probably review your trigonometry. But one thing you should remember is that the trigonometric functions are related by several useful identities. In particular, there are what are called the Pythagorean identities. For all x, sine squared of x plus cosine squared of x is equal to one, and tangent squared of x plus one is equal to secant squared of x. And there are also identities involving cosecant and cotangent, but unless you have too much free time on your hands, those identities are probably not worth learning. So let's consider the integral of cosine cubed of x. So what do we know? Well, we do know the Pythagorean identity, and that means we can say something about cosine squared of x, and so this integral is the same as the integral of one minus sine squared of x times cosine x. And we can fall back on a u-substitution. If we let u equal sine of x, then cosine x dx is going to be du, and so. And that gives us anti-derivative u minus u cubed over three, putting everything back where we found it, and including our constant of integration. How about the integral of tangent squared x secant to the fourth x dx? We'll use our Pythagorean identity, and what the heck, we can solve this for tangent squared of x. And so we can substitute that in, but it's not entirely clear what we should do at this point. While we might get this to work, let's try something else and see if we can do this more easily. So let's flip that Pythagorean identity around, and use secant squared equals tangent squared x plus one. And since this is a secant to the fourth, that's really secant squared squared, so we'll make that substitution. And again, it's not entirely clear how we want to proceed from this point. Well, third time's a charm. We remember that the derivative of tangent is secant squared. And so one possibility is maybe let's keep one of these secant squareds and convert everything else to tangent using our Pythagorean identity. And now we can try a u-substitution with u equals tangent x, and du is secant squared of x dx. Now we can integrate, and putting everything back where we found it gives us our definite integral.