Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. 3 basic trig substitutions. We're going to let sqrt X ^2 + a ^2 equal a secant Theta. If we thought about drawing a triangle, we could see that secant Theta would just be sqrt X ^2 + a ^2 / A. So if we drew our triangle, our hypotenuse would be that sqrt a ^2 + X ^2 and our adjacent side would be A, which would leave our unknown side to be X or our third side to be plain old XA. In this case, is going to be some constant. So we could have a relationship that says the tangent of Theta is X / a, or we could think of that as X equaling a tangent Theta. So if we wanted to solve for Theta, we would have tangent inverse of X / a equaling Theta. We're going to restrict that remembering about our tangent graph. If we think about our tangent graph, when we do the inverse, we want it to be 1 to 1, so we want all the values in between negative π halves and Pi halves. That's going to be an important restriction. So we know that secant Theta is greater than 0 in the restriction of negative π halves less than Theta less than π halves because secants greater than 0 in this restriction. Then if we have sqrt a ^2 + X ^2, we could think of that as sqrt a ^2 plus instead of X ^2. Let's put in a tangent Theta and square it. Then we can see that there's an A ^2 in both those terms, and we get one plus tangent squared Theta, which is a Pythagorean identity. And that's really just secant squared Theta and sqrt a ^2 is A. The square root of secant squared Theta is going to be secant Theta because secant Theta is positive in those restricted boundaries. Negative Π halves to Pi halves and the A is going to be a distance, so the A is always going to be positive also. So if we look at our next one, we're going to think about if we let X equal a sine Theta. So sine Theta is X / A. If we draw our right triangle, we get X / A. Our third side would then be sqrt a ^2 -, X ^2. So if we solve for Theta, we get sine inverse of X / a equal Theta with sine. Once again, think about our graph and our restrictions for sine are going to be negative π halves to Pi halves because it needs to be 1 to one. We need to cover every value exactly once. So negative π halves less than or equal to Theta less than or equal to π halves with sqrt a ^2 -, X ^2 instead of X ^2. We're going to put in what it equals, which is a sine Theta squared. Factoring out the a ^2, we get one minus sine squared Theta which is cosine squared Theta and square root of that is going to be a square root of cosine squared Theta. Cosine is positive in the restrictions for Theta, so we get a cosine Theta. We're going to use these in integrations for substitutions in a few moments. So our last one is going to be X equal a secant Theta. We could think of X / a equals secant Theta, so hypotenuse over adjacent. That makes our third side be sqrt X ^2 -, a ^2. So we could see secant inverse of X / a is Theta with our secant inverse. If we think about that graph, we have a graph that looks something like this with an asymptote and a graph that looks something like this. So we're going to be looking at 0 to pie halves, not including our pie halves because the asymptote union pie halves to pie. If we have instead of X ^2, we put in what it equals, a secant Theta squared minus a ^2. We factor out this a ^2 secant squared Theta -1 sqrt a ^2 tangent squared Theta because this is a Pythagorean identity. It's going to equal a tangent Theta if we're on zero to π halves, because tangent Theta in the first quadrant is always positive. But it's going to equal the opposite of a tangent Theta between π halves and π because tangent in the second quadrant is negative. Another way to write it as we could think of it as a equal or equal a, the absolute value of tangent Theta. And that'll take into account what's happening because if we have a tangent in between π halves and π that's negative, we'd have the opposite of a negative, which would really make it a positive A times tangent Theta. So if we have sqrt X ^2 -, a ^2, just a refresher, we want to think about the square root X -, a * X + A. This X + a is always going to be positive. So in order to get this to be a real number underneath, we need the X -, A to also be positive. So and it could technically equal 0, but we're talking distances, so we really want it to be positive. So X is going to be greater than a. Or we could divide each side by A and get X / a greater than 1A is a distance, so we know we don't have to worry about flipping the inequality side by dividing by a negative. So X = a tangent Theta when X / a is greater than one. So that's another way of coming back and thinking about this here. If we think about an example, if we have 3D Y integral square root 1 + 9 Y squared, the first thing we're going to do is think about these as squares. So 1 ^2 + 3 Y squared. And we want to think about this three Y being which one of those 3 trig identities. And we're going to say tangent Theta because when we draw our triangle, we want the three Y to be opposite the one to be adjacent so that we can have sqrt 1 + 9 Y squared being on the hypotenuse. So what we're really doing is we're looking at this piece here to figure out where we're going to put our sides in order to know which of the three trig substitutions we're going to use SO3Y over one. This is a right triangle is going to give us the tangent Theta and that third side square root nine y ^2 + 1 is what's in the denominator. So if we know 3 Y is tangent Theta, we can have Y being 1/3 tangent Theta. 3D Y is going to be secant squared Theta D Theta. So now when we integrate this 3D Y, we're going to substitute secant squared Theta D Theta over sqrt 1 plus instead of the three Y quantity squared, we're going to put in tangent squared Theta. But one plus tangent squared is secant squared and the square root of the square is going to cancel. So we're going to get a secant Theta to cancel top and bottom. And we know that the integral of secant Theta D Theta is just the natural log of the absolute value of secant Theta plus tangent Theta plus C. Now if we go back to that triangle, our secant Theta is the square root of nine y ^2 + 1 / 1 and our tangent Theta was three y / 1. So our final answer is going to be the natural log of the absolute value of square root nine y ^2 + 1 + 3 Y plus C We actually know that this is going to be a positive, and we know that Y is a distance, so it's actually going to also be a positive. So we could use parentheses there if we prefer, looking at another example. So on this example, we have one sqrt 1 -, 9 T squared. So when I set up my triangle, I want to have my one on my hypotenuse because I'm going to have one minus. If we're thinking about a ^2 + b ^2 = C ^2 and we're going to have something about something minus, we know that that first thing is always going to be the hypotenuse minus. In this case, we're going to put 3T on the opposite side so that we have sine of 3T being 3T or sine of Theta being 3T over 1. So if we have sine of Theta being 3T over one, that makes our third side our adjacent side sqrt 1 - 9 T squared. So if we have sine of Theta equaling 3T3DT is going to equal cosine Theta D Theta or DT is 1/3 cosine Theta D Theta. So now we're going to do some substitution. Here's my 1/3, here's my cosine Theta D Theta. The square root 1 minus instead of nine t ^2 that's going to get substituted as sine squared Theta. Well, we know that one minus sine squared Theta is really cosine squared Theta, and the square root and a square are going to cancel. So we're going to get 1/3 cosine squared Theta, D Theta. Now we don't know how to integrate this, but we have some rules that say we can change a cosine squared Theta by the power reducing formula into one plus cosine twice the angle over 2. So now we get 1/3 the integral of one plus cosine 2 Theta over 2. Integrating each of those pieces, I'm going to pull the two out in front and the 1/2 in front. So 1/6 Theta plus 1/2 sine 2 Theta plus C. Now we don't want to have it in terms of Theta because the original was in terms of T So we're going to look here and we're going to realize that Theta is really sine inverse of 3T. So we're going to have 1/6 sine inverse 3T plus 1/2. Our triangle is Theta, not 2 Theta. So we're going to use our double angle identity that says 2 sine Theta, cosine Theta and then we have a + C So let's see. Looks like I forgot a parenthesis there. So 1/2 * 2 is going to cancel sine of this triangles 3T over one cosine of sqrt 1 minus nine t ^2 / 1 + C. So we could actually check this. We could check all of these if we wanted. So if we know that this is what we just evaluated the integral for, if we actually took the derivative, it should come out the same. So 1/6 sine inverse of 3T. The derivative of the three T is 3 over the square root 1 - 3 T squared, in this case nine t ^2 plus. Now we're going to do the product rule. The derivative of 3T is going to be 3 times the square root of the 1 - 9 T squared plus the 3T times the derivative of this. Using the chain rule, we have -18 T over 2 times sqrt 1 - 9 T squared plus the derivative of constants really just zero. So technically down here we have a + 0. Now, if we simplify this up, we'd get 3 + 3. If I get a common denominator here, I'm going to multiply top and bottom by sqrt 1 - 90 ^2. So 3 * 1 - 90 ^2 - 27 T squared. That doesn't seem right. 3 * 8 is 4/24 and 54. Oh, except I reduced two goes into a negative eighteen. 9 * 9 * 3 is 27 OK 27 T squared. So we'd have 1/6 of 3 + 3 - 20 seven t ^2 - 27 T squared, or 1/6 of 6 - 54 T squared's over sqrt 1 - 9 T squared, which would give us 1 minus nine t ^2 / sqrt 1 - 9 T squared. When I simplify that, I get sqrt 1 - 9 T squared. If we look at another example, we're going to have five the integral of five DX over the square root 20 five X ^2 - 9 where X is greater than 3/5. That X is greater than 3/5 is important so that the inside of the square root is positive. So in this case, we're going to let 5X equal 3 secant Theta because once again, on this triangle, I want the first term because it's a minus to be on the hypotenuse SO5X over three, which would give us our other side P to be sqrt 20 five X ^2 -, 9. So we know that five X / 3 is secant Theta. If we come back up here, five X / 3 is secant Theta. So if we take the derivative of each side, we get 5 DX equaling 3 secant Theta tangent Theta D Theta going to take out the five DX going to substitute what it equals square root. Instead of 25 X squared, we're going to put in nine secant squared Theta -9 pulling out sqrt 9, which is 3. So those threes will cancel. Secant squared Theta -1 is a trig identity, it's the tangent squared. And the square root of the tangent squared is going to leave us tangent. So those tangents will cancel. So all of this really reduces to the integral of secant Theta D Theta. Well, we know that the integral of secant Theta D Theta is the natural log of the absolute value of secant Theta plus tangent Theta plus C or the natural log. Looking at our triangle, instead of secant Theta, we're going to have 5X over 3 plus instead of tangent Theta, we're going to have sqrt 25 X squared -9 / 3, and that's plus C Looking at another example, we have the integral of DX over X ^2 sqrt X ^2 + 1, so I'm going to let X be tangent Theta. When I draw my triangle, I get X / 1, so my third side's sqrt X ^2 + 1. When I take the derivative of each side, I get DX equaling secant squared Theta D Theta. So we have secant squared Theta D Theta for the DX instead of the X ^2 we're going to put in tangent squared Theta, square root tangent squared Theta, plus one trig identity again. So the square root of secant squared Theta is going to give me secant Theta canceling top and bottom. So I'm going to get a secant Theta D Theta over tangent squared Theta. If I change this all into cosines and sines just so we can see what's happening, we get the integral of one over cosine Theta times cosine squared Theta over sine squared Theta D Theta, which reduces to cosine Theta over sine squared Theta D Theta. Now if we did AU sub and we let U equal our sine Theta, we'd have DU equal cosine Theta D Theta. So we'd end up with the integral of 1 / U ^2 DU, which would actually evaluate to -1 / U + C, which is -1 over sine Theta plus C, which is really negative cosecant Theta plus C. And we look at our triangle because we need this in terms of the XS that we started with. Cosecantness is going to be sqrt X ^2 + 1 / X + C In this example, we're going to let our tangent Theta be the LNY so that when we take the derivative we have 1 / y DY equaling secant squared Theta D Theta. So when we do our substitution, we're going to have secant squared Theta D Theta in the numerator over sqrt 1 plus tangent squared Theta. One plus tangent squared Theta we know is secant Theta and then it's secant squared Theta and the square root gives us secant Theta, so that reduces to secant Theta D Theta. I did not change our bounds because it will be difficult to change the bounds in this situation. So we're going to substitute them back in the end. The natural log of secant Theta plus tangent Theta from A to B. When we look at our triangle, if we knew that the natural log of Y was tangent Theta, the opposite sides natural log of Y adjacent sides 1, sqrt 1 plus natural log of y ^2 is the hypotenuse. So secant Theta is that sqrt 1 plus natural log of y ^2 plus tangent Theta which is just L&Y. Putting in our bounds. Going back to our original bounds of one to E, we get the natural log of sqrt 1 plus the natural log of E. Natural log of E is 1, so 1 + 1 is just two plus the natural log of E which is one again minus the natural log of sqrt 1 plus the natural log of one square. Natural log of one is 0, so we get sqrt 1 which is just one, and then the natural log of one which is 0, so we get 1 + 0 and then the natural log of one is 0 again. So that last term actually just goes to 0. So we end up with the natural log of sqrt 2 + 1 as our final answer. If we look at one more example, we get X ^2 + 1 quantity squared DYDX equaling sqrt X ^2 + 1, Y of 0 equal 1. This is an initial value problem, and so we're going to get our YS on one side and everything else on the other, and then we're going to integrate. So if I have the integral of DY equaling the integral of sqrt X ^2 + 1 / X ^2 + 1 ^2 DX, if I let X be tangent Theta. So we have X over 1 sqrt X ^2 + 1 on our third side of our right triangle. DX is secant squared Theta D Theta. So we're going to take out this DX. We're going to put in secant squared Theta D Theta. We're going to have the square root of tangent squared Theta plus one over tangent squared Theta plus one quantity squared. This inside quantity is just secant squared. So secant squared squared is secant to the fourth Theta, the square root of secant squared Theta plus one. Well, the square root of tangent squared Theta plus one, which is really secant squared Theta gives us a single secant times another secant squared or secant cubed Theta. That really simplifies into just cosine Theta D Theta. And we know that cosine Theta D Theta, when we do the integral, that's just going to give us Y equaling sine Theta plus C And the initial bounds were given as Y of 0 equal 1. So sine of 0 is 0. So C is our 10. We're not going to do that yet, sorry. This was in terms of XS. This is Y of X. So we need to substitute back from our sine Theta. So our sine Theta if we look at our triangle here. Our sine Theta is going to be X / sqrt X ^2 + 1 + C Now that zero is going in for our X's everywhere and our C is going to end up being 1. So when our C is one, we get Y equal X square root X ^2 + 1 X over square root X ^2 + 1 + 1. Thank you and have a wonderful day.