 This lesson is on trigonometric substitution. Let's look at a few integrations and see how certain ones lend themselves to trigonometric substitution. The first one reads the integral of dx over x squared times the square root of 9 minus x squared. When you look at that square root, it should remind you of arc sine 1 minus x squared. So that in your mind means that it needs some sort of trig substitution. And sure enough, we would use trig substitution on this particular problem. On the second one, which reads the integral of dx over the square root of 4x squared plus 1, again, it should remind you of an inverse function. And this time, it's the derivative of pretty much an arc tangent. Again, lends itself to trig substitution. The third one is the integral of dx over the square root of x squared plus 1 to the 3 halves. Again, x squared plus 1 should make you think some sort of arc tangent derivative, true. And it lends itself to trig substitution. And the fourth one, the integral of dx over the square root of x squared minus 6x plus 10, doesn't lead to any trig substitution that you might know with your inverse functions. And this one is not for trig substitution. It actually is completing the square type of problem. So recognizing these becomes very, very important because it helps you in the way you begin. So let's look at what you do when you actually begin to make substitutions for your trig. As you look at these, you realize you have three types of trig substitution. You have the sine, you have the tangent, and you have the secant. Let's work on the sine one first. Make a right triangle. This is always a good way to begin. Now set this right triangle so that you have u here, a function in x, and an a there, some sort of a constant. So what we have done is set it up so that sine theta is equal to u over a. Our u is equal to a sine theta. That would make our du equal to a cosine theta d theta, and we need all of this for our substitution. What happens to the third side? Well, the third side on our triangle becomes a squared minus u squared, the one you recognize in order to make this substitution. And a squared minus u squared in that radical becomes a squared minus a squared sine squared theta. And that becomes, take out the a squared where left with one minus sine squared under a radical becomes a cosine theta. Nice, simple substitution. Let's go on to another type of substitution. Set up our triangle this way with the u here and the a there. And you can see that tangent theta now is u over a, or u is equal to a tan theta. That makes du equal to a secant squared theta d theta. The radical for the hypotenuse is u squared plus a squared. So that radical u squared plus a squared becomes the radical of u squared, which is a squared tangent squared theta plus a squared. Factor out the a squared and we have tangent squared theta plus one, which is secant squared theta. So this becomes a secant theta. Now some people just memorize these formulas as they are. I tend to derive them for every single problem I do and we'll derive them again once I do official problems. Let's go to the last type, which would set up the triangle again with our theta, our right angle, with u here and a there. And this time we will say secant theta is equal to u over a. Again, u is equal to a secant theta, du is equal to a secant theta, tan theta, d theta. And then the radical this time would be u squared minus a squared. So that radical u squared minus a squared will be equal to a squared secant squared theta minus a squared. And if we factor out the a squared we get secant squared theta minus one. And all of that will lead to a tan theta. So those are the formulas that we put together in order to do trig substitution. So let's go on and do some problems. Our first problem reads x squared and then square root of 9 minus x squared. So again, this one looks like a sign and we will set it up in that manner. So make the right triangle with your theta and remember it's u over a. So in this case u is x, a is 3. And so we say sine theta is equal to x over 3 or 3 sine theta equals x. So that makes dx equal to 3 cosine theta d theta. And then our square root of 9 minus x squared becomes the square root of 9 minus 9 sine squared theta. So we would say before was in this case the a will be 3 cosine theta. So now we can do all that substitution into our integral. We have the integral dx is 3 cosine theta d theta. The x squared is 9 sine squared theta and the radical is 3 cosine theta. Everything has been substituted in and remember I even substituted in for dx. So the 3 and the cosine go out. So I have d theta over 9 sine squared theta. Well that becomes sine squared can move up to cosecant squared over 9 or pull the 1 ninth out, which might be a little bit better to do. And this becomes negative 1 ninth cotangent theta plus c. Substituting in for cotangent because we have to get this back into x's negative 1 ninth, cotangent in our case is remember this is the square root of 9 minus x squared. So cotangent will be the square root of 9 minus x squared over x. And that of course would be plus some constant. So simple if you practice it of course, but it's a really neat way to solve some of these problems. Let's go on and do another problem. This one reads the integral of dx over the square root of 4x squared plus 1. Again looks like an inverse trig function so we are going to use tangent on it because it looks like the inverse trig function for tangent. So if we set up our triangle, our u is 2x, our a is 1. So this becomes the square root of 4x squared plus 1. So if we make tan theta is equal to 2x, or x is equal to 1 half tangent theta. Take the derivative of this, it's dx is equal to 1 half secant squared theta d theta. Doing the square root for x squared plus 1. We will square our 1 half tangent theta. So the 1 half becomes 1 fourth. That'll cancel out the 4. And then we will have tangent squared theta plus 1. And that all equals under the radical of course secant theta. Substituting back into our integral we have dx, which is 1 half secant squared theta d theta, over the square root which is secant theta. This gives us 1 half integral of secant theta d theta. And the integral of secant theta is 1 half ln of the absolute value of secant theta plus tan theta plus under all absolute value plus some sort of a constant. We need to finish this one off by putting it in x's. So we have 1 half ln and secant theta seems to be the square root of 4x squared plus 1. So it's ln of the absolute value of the square root of 4x squared plus 1. Our tangent is 2x. So it's plus 2x and then plus our constant. Again, this went a little bit different than the last one because we had to substitute with the tangent. Let's go on and do one more. Integral of x over the square root of x squared minus 8 dx. This one is neither sine nor tangent, it's actually secant. And when we set it up, here's our theta. We will have x here and the square root of 8 here. So our radical will be x squared minus 8. In this case, we will say secant theta is equal to x over the square root of 8 or x is equal to the square root of 8 secant theta. That makes dx equal to the square root of 8 secant theta, tan theta, d theta. And our radical square root of x squared minus 8 will equal the square root of 8 secant squared theta minus 8 factor out the 8 and we get secant squared theta minus 1, which is tangent theta. So we get square root of 8 tangent theta. Put this all back into our integral and we get integral of x, which is the square root of 8 secant theta, dx, which is the square root of 8 secant theta, tan theta, d theta over the square root of x squared minus 8, which is the square root of 8 tan theta. The square root of 8, at least one of them goes out, a tan theta goes out, and we are left with square root of 8 in that numerator and secant squared theta, d theta. Anti-derivative of secant squared is tangent plus c. Again, we need to put in our x's. So let's go back to how tangent is defined. And tangent theta is equal to the square root of x squared minus 8 over the square root of 8. Those cancel out and we get a plus c. Trigonometric substitution is a very interesting way to do integration. This concludes your lesson on trigonometric substitution.