Hello wonderful mathematics people, this is Anna Cox from Kellogg Community College. Indefinite integrals. If U is any differentiable function, then the integral of U to the NDU equals U to the n + 1 / n + 1 + C where N is not equal to -1. So if we thought of letting U equals some function G of X and G of X is a differentiable function whose range is an interval I and F is continuous on I, then the integral of the composition function F of G of X times the derivative of G of X DX equals FUDU the integral FUDU the thought process. Think about letting U be the inside function of that composition of G of X. Then if we take the differential of each side, DU is just the derivative of G of X times the differential of DX. So DU equal G prime XDX. So I'm reviewing for trig functions. If we know cosine 2X, that's really cosine X + X. So we could think of that as cosine X times cosine X minus sine X times sine X, or cosine squared X minus sine squared X if I didn't want to have a cosine squared X in there. If I wanted this only in terms of sine squared XS, I'd replace the cosine squared X using the Pythagorean identity of 1 minus sine squared X, and I'd get 1 -, 2 sine squared X. Now if we wanted to solve for sine squared X, we would take everything that didn't have the sine squared X to one side, and then we would divide by the coefficient. So sine squared X would equal 1 minus cosine the quantity 2X over 2. If we wanted to solve for just sine X, we would take the square root of each side. So the sine squared X is called a power reducing formula, and the sine X equaling positive negative square root 1 minus cosine 2X over 2 is frequently thought of as the half angle formula. If we went back to the equation cosine squared X minus sine squared X and did the same steps but solving for cosine squared X this time, we'd get cosine squared X equaling one plus cosine 2X over 2 or cosine X equal positive negative the square root 1 plus cosine 2X over 2. If we look at an example secant the integral secant 2T times tangent 2 TDT. We're going to do substitution here. So let's let U equal to T. If U equal to T, then DU is going to equal to DT by taking the differential of each side. We know that we have a DT up here. So if we solved for this DT, we could say 1/2 DU equal DT. So now I'm going to substitute secant. Instead of two T, I'm going to put in U tangent Instead of two T I'm going to put in. UI can't have a DT because I don't have AT variable anymore. I have AU variable. So instead of the DT, I'm going to put in 1/2 DU. Now the 1/2 can go out in front because it's a constant. Constants can move in and outside of the integral freely, so see if it's multiplication. So secant U tangent U du. What gives us out a derivative of secant nu tangent U? So we're going to get 1/2 secant square secant U + C. Now the original was not in terms of secants U's, the original was in terms of T's. So we need to figure out how to get secant U to be in terms of T's. And we're just going to do a straight substitution back instead of U. It's going to be two t + C If we thought about the derivative of this, the derivative of secant is secant tangent. The derivative of the chain by the chain rule of 2T is two 2 * 1/2 would make it cancel. So there would be A1 coefficient and the derivative of C is just zero. Looking at some more examples, we have the integral of nine R-squared Dr. over sqrt 1 -, r ^3. Well, we're going to let substitution again. This time we're going to let U equal the inside of that square root. So if I have U equaling 1 -, r ^3, we're going to get DU equaling -3 R-squared Dr. So we usually don't cross variables over the =. So I want to leave all the Rs and DR's on one side, but I can move that -3 and make it -1 third DU. There are truly lots of ways to do this. This is one method that I think is easy or easier. So I'm going to move the coefficient to the other side. When we look up here, we're going to have nine. This R-squared Dr. is right there, and that R-squared Dr. is really just negative 1/3 DU, and then that's over the square root. Instead of 1 -, r ^3, we're going to say square to unit 9 * -1 third is going to be -3. And because it's a multiplication of a constant, I can move it inside and outside the integral at will. So now we're going to add 1 to the exponent and divide by the new exponent. Remember we still had that -3 in front and we're going to have plus C. So when we simplify this, we get -6 U to the 1/2 + C. Our original didn't have any U's in it, so we need to do a substitution back negative six 1 -, r ^3 to the 1/2 + C. Or we could think of that as -6 square roots of 1 -, r ^3 C To check this, let's take the derivative in terms of R of that function. So if we wanted to check it, the derivative of a square root is 1 / 2, the square root of the inside. But then we need to take the derivative of the inside. The derivative of one is 0. The derivative of negative r ^3 is -3 R-squared, and then the derivative of that constant at the end of 0 so -6 * -3 positive 18 / 2 is nine r ^2 / sqrt 1 - r ^3, which is what we started with doing another one. This one is more complex, and we need to figure out what we want our U to be. If we thought about letting U be sine of X / 3, we'd end up with AU to the fifth power, and we could use our power rule. If we thought about letting U be the cosine X ^3, that's just plain U, and it's not as complex as it could be. We always want to go with the most complex substitution. So if I let U equal the sine of X / 3, DU is going to equal 1/3 cosine X / 3 DX. I can move that 1/3 to the other side to get three DU equal in cosine X / 3 DX. So now we're going to have the integral sine to the fifth power. Instead of X / 3, we're going to call it U. This cosine X / 3 DX is right here, and that whole thing equals 3 du. I can pull the three out in front because it's multiplication way back here. You guys are all screaming wait a second. And you did it wrong. Hopefully. Sine X / 3 is U. So this whole thing turns into U to the fifth, because sine X / 3 to the fifth power. So if sine X / 3 is U, we have U to the 5th, so U to the fifth three DU. So the three's going to come out in front U to the fifth DU. Add 1 to the exponent. Divide by the new exponent. Remember there was a three in front and we're going to have a plus constant. So we're going to get 1/2 U to the 6 + C, But the original problem didn't have U's in it, so we're going to substitute and we're going to get 1/2 sine the 6th power of X / 3 + C So to check, we're just going to take the derivative in terms of X of that function. So the derivative of sine 6X over three, we're going to bring down the power to 1 less power. Then we have to multiply that times the derivative of sine which is cosine, times the derivative of X / 3 which is 1/3 and the derivative of the constant 0. So 1/2 * 6 * 1/3 is going to give us A1 coefficient and we're going to get sine to the 5th X / 3, cosine X / 3. So our next example we have X ^3 sqrt X ^2 + 1. We're going to let U equal X ^2 + 1, so DU is going to equal 2 XDX. If we if we take that two to the other side, multiply by 1/2, we get 1/2 DU equaling XDX. So now we have an X ^3. So that's really an X ^2 * X sqrt X ^2 + 1 DX. Well, we know that this X DX is really going to be 1/2 DU. We know sqrt X ^2 + 1 is going to be the square root of U. But now we have to figure out what do we do with this X ^2 that's still left? Well, we're going to come back to this original equation of our substitution. If we know that U = X ^2 + 1, then we know that X ^2 is really U - 1. So instead of that X ^2, we're going to put in U - 1, so we're going to have 1/2. We're going to multiply things out. So we're going to get U to the three halves minus U to the one half DU, adding one to the exponent and dividing by the new exponent, adding one to the exponent, and dividing by the new exponent plus C So we're going to get. If we flip that and multiply, we're going to get 1/5 U to the five halves -1 third U to the three halves plus C. So the original wasn't in terms of use. The original was in terms of X's. So one 5th X ^2 + 1 to the five halves -1 third X ^2 + 1 to the three halves plus C. If we wanted to check, we would take the derivative and terms of X of that whole thing, remembering we have to use the chain rule over and over until we've done all the pieces. So bringing down the power to 1 less power and then times the derivative of the inside, the next term we're going to bring down the power to 1 less power, times the derivative of the inside. And then the derivative of the C is just zero. So we're going to get one half. Oh, 1/2 * 2 is just going to give us ** squared plus one to the three halves. Here we're going to get the one thirds to cancel, the twos to cancel. So we're going to get minus X * X ^2 + 1 to the one half. So what was our original X ^3 sqrt X ^2 + 1? So if we factor out a sqrt X ^2 + 1, or we could think of that as X ^2 + 1 to the one half. That would leave me an X times an X ^2 + 1 minus an ** squared plus one to the one half X ^3 + X -, X the plus X -, X cancel. So we'd have sqrt X ^2 + 1 * X ^3, which is indeed what we started with. Thank you and have a wonderful day.