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Video transcript

let's see if we can evaluate the indefinite integral one over nine plus x squared DX and we know that if you have the pattern a squared minus x squared it could be a good idea to make the substitution X is equal to a sine theta but we don't see that pattern over here what instead what we see is a squared plus x squared and in this context it tends to be a good idea it's not always going to work but it never hurts to try out there's a little bit of an art here to try out X is equal to a tangent theta now you might say Sal why is that well let's make that substitution and see how this thing would simplify this thing would become a squared plus a squared tangent squared theta which is a squared times 1 plus tangent squared theta and this right over here this right over here we could reprove it actually let me just reprove it for you this is going to become a squared times this is 1 can be written as cosine squared theta over cosine squared theta tangent squared is sine squared theta over cosine squared theta and this is why I picked cosine squared as the denominator so that I can add the two and this is going to become a squared times cosine squared theta plus sine squared theta all of that over cosine squared theta this numerator from the unit circle definition of trig functions becomes 1 so this is 1 over cosine squared theta and everything simplifies to a squared secant squared theta which might simplify things so let's see if we have what's going on over here we could rewrite this so 9 plus x squared 9 plus x squared you could rewrite as 3 squared plus x squared in this case in this case a would be equal to 3 so we'd want to make the substitution X is equal to 3 3 tangent of theta and if we wanted to solve for X you can divide both sides by 3 because we're later going have to undo the substitution x over 3 is equal to tangent theta or theta is equal to arctangent or inverse tangent of x over three now we're also going to have to figure out what DX is we're also gonna have to figure out what that is so let's take the derivative or we'll write a differential form DX is equal to three derivative of tangent theta with respect to theta is secant squared theta D theta so now it looks like we're armed with all the things necessary to rewrite this entire integral it's going to be equal to the indefinite integral you're going to have DX here which is equal to 3 secant squared theta D theta that's the DX and it's all of that it's going to be over reducing this all of that is going to be over this business right over here our a squared plus our x squared now we already know what that's going to simplify to our a squared plus our x squared is going to since we made this substitution X is equal to three tangent theta X is equal to three tangent theta is going to simplify to a squared secant squared theta so this is going to simplify to 9 9 secant squared theta and you could read you could essentially go through this logic you're going to get 9 plus 9 tangent squared theta factor out a 9 you get 9 plus 9 times 1 plus tan squared theta and so that's going to be 9 times secant squared theta exactly what we have here lucky for us we have the secant square it's cancelling out the secant squared is cancelling out you have 3 over 9 this whole thing can be rewritten as 1/3 that's just the 3 over 9 times the indefinite integral of just D theta which is equal to 1/3 theta plus C and now we just have to put things in terms of X and we see theta is equal to arctangent of X over 3 so this is going to be equal to 1/3 arc tangent arc tangent of x over 3 plus C and we are done so now we know how to deal cases where we might want to where we see something like an a squared minus an x squared and an a squared plus an X it won't always work but it might be a useful it'll definitely allow you to do this thing it might not always make the integral solvable but it's not a bad thing to try when you trip when you when it looks like u substitution isn't working then you could look for these patterns and try some trig substitution