3.5.1

Instructor: Example one: use the table formula here to evaluate the antiderivative of the square root of 16 minus e to the 2x divided by e to the x dx. Now, the first thing you want to observe is that this integral somewhat resembles this integral. Somewhat. So if we define a and u to be certain values, then this might actually match. Now something you may also notice is that this has a u squared and a u squared, whereas this has what, if u was e to the x, then that's a u squared, but that's just u that's actually going to work itself out once we do our substitutions. 

So let's begin. We say a equals 4 because that would be 16 squared. And u must be e to the x, because e to the x squared. If you think of that, e to the x quantity squared, multiply in the exponents, that's e to the 2x. So there's our choice there. So now we've also got to make a substituent for du. So du would be e to the x dx, which does not, well it almost matches what we've got here. Almost matches. So I'm going to make an adjustment here. This is du equals u dx, and I would prefer to write this as du over u equals dx. So I'm going to make a substituent for dx that's in terms of u. 

So rewriting that integral, I'm going to have the integral of the square root of a squared-- and again, I want this to look like that formula-- minus u squared, so far so good, e to the x. That's not good. e to the x is just u. That's not a u squared. However, once I throw this in here, I'm going to replace dx with du over u. If you notice, that will be the integral, square root of a squared minus u squared over u squared du. And we look exactly like the formula. 

So that means the answer to this integral problem will be negative square root of a squared minus u squared over u minus arcsine of u over a plus C. Now replacing a's and u's with our known values here, that is minus square root of 16 minus e to the 2x, using the fact talked about earlier, divided by e to the x, that's our u, minus arcsine of e to the x over 4 plus C.