00:01
For problem 35, we're given an integral that we need to solve, but we're going to have to use u substitution to do it.
00:08
So looking at this here, the derivative of cosine is going to lead us back to sign.
00:13
So let's try using u equals cosine of 2t plus 1, right? and so if we take the derivative that du is going to equal, well, the derivative of cosine, right? that's negative sign, so negative sine times 2t plus 1 dt, but then we have the chain rule here, so we need to also multiply it by the derivative of the inside, which is just 2.
00:40
So as we kind of predicted, this is starting to come together quite nicely, right? we have our cosine right here that can be squared and substituted in, and now we have a sign as well, right? sign and this d t that is right here as well so um the only thing we have left is this negative two so let's flip that over to the other side right so if we solve for just sign of two t plus one d t this is going to equal negative one half times d u so let's substitute all of us in we'll pull out that negative one half up front and we get the integral right of one over u squared d u now this is starting to look like something we can actually work with so we get negative one half there's the integral view to the negative 2, right? and now we can integrate and we get negative 1 half.
01:30
We add one to that power.
01:31
That becomes u to the negative 1 and then divide...