5

Step And now we finally have36 e58 sin(68) d8 = I = 2525 C, where C =Putting all of this together and incorporating the constant of integration, C, we have| ese sin...

Question

Step And now we finally have36 e58 sin(68) d8 = I = 2525 C, where C =Putting all of this together and incorporating the constant of integration, C, we have| ese sin(6e) 40

Step And now we finally have 36 e58 sin(68) d8 = I = 25 25 C, where C = Putting all of this together and incorporating the constant of integration, C, we have | ese sin(6e) 40



Answers

$33-38$ First make a substitution and then use integration by parts
to evaluate the integral.
$$\int t^{3} e^{-t^{2}} d t$$

Look at this question. You're asked to find the value of this integral by making some sort of substitute. So we're gonna solve this question by using some sort of substitution. So the first sassy is I realized, OK, there's an E to the X. You can essentially turn into you. So I'm gonna say you equal to each the X because I think this is the most logical way to go about this and the reason I think that is because there's an e to the X on top and on bottom. And so when you do d you over dx um, just to find e of X as well, it's gonna be the same thing. So it really becomes a relatively simple question. Um, so let's set you equal ttx so it's gonna be you squared over. Ah, let's see. Do you? Because do you over e to the X is equal to Devi. We're sorry is equal to d X and so it's gonna be the integral of you because that's each of the X Over you square plus three u plus two. Okay, so this question actually looks like a very familiar question. This looks like a question where you would use partial fractions. Some Let's go here and solve for the partial fraction for this. So what I see is I see a U plus to any U plus one. Um, so I'm gonna do you plus one, and you plus two. I'm just gonna use partial fraction decomposition for this question. So when you have a you plus one and that you plus two, what do you do? Your in a set? A you plus b you equal to you and you're gonna set to a plus B equal to zero. So you have from this as you know, that a plus vehicles one. So I'm gonna recent, if I like that and then I'm going to do some relatively simple subtraction. So I'm going to subtract the bottom from the top. So let's see, when I have here is I get negative. A many cross out the bees and it becomes equals one. So a equals negative one. Now that have this, I know that a plus B equals one. So I know be is equal to two. It's not gonna have a and B I can go ahead and relatively simply solve this question. So now what a way to have I have the integral of one over the negative interval of one over U plus one plus the integral of to over U Plus two. So this becomes too simple longer than so. I'm gonna rewrite this so this becomes negative Ln of you plus one. And this becomes too Ln u plus two. Okay, So I know by longer than rules that when you have to log rhythms with different bases, you can set one over the other, especially when there's a negative sign, because the negative sign means that essentially the Ln of you plus one is being divided from the overall natural longer them. So let's rewrite this as Ln of you plus two squared because that's how that would work, right, because it be the entire term squared over. Sorry, You're just subtracting from the Ln part Ln of use plus two squared over U plus one. And after this, we just re substitute and you can remember from the original question we had u equals X. So now that we've solved for everything else, it just comes Ln of you to the X plus two whole squared. So all of this squared over e to the X plus one. Okay, so and then because it's an indefinite internal your toe add C. So the final answer for this question is Ln of e to the X close to whole squared over eat the X plus one plus c. Okay, so I'm gonna go back and explain how this was done. So the first thing I noticed is there's a young, the top in the bottom and it looked like a relatively simple partial fractions question. So I said, you equal to e to the X. And then I found a partial fraction for this which came from this what? Which is deep product of the use of So so now that I have the use of I used the partial fraction and I saw for A and B after I'd solve for N b. I found to Logar thems and after I found the two large items, I realized OK, they could be simplified together using longer rhythmic rules, especially this rule that the uh coefficient oven Ln becomes the exponents and the other will saying and negative Ln becomes a quotient in the overall span of ah, natural logger room. So now I have Ln of you plus two whole squared over U Plus one and then you finally plug in e squared. They're e to the X into you. Define the final answer. It's the final answer to this question is Ln of e to the X plus two whole squared over E to the X plus one all plus seat. That's a final answer for this question. I hope that helps for using U sub to help find the partial fraction decomposition of an indefinite integral.

Five. We need toe. Evaluate this integral off the form a sign off a function which is a sign off Lexa function you x plus c using the appropriate choice of the U. S. So when it comes, assign, it means that it can be anyone off the trick. No metric function, whether it's sine or cosine. So that totally depends on how choice of the integral is. Uh, let's make a substitution over here as X rays to the poor four. As let's say, t because if we do that and we differentiate both sides, we have four x cubed DX, that which is the differentiation of extra stop all four. This comes as GT and if we divide both sides before we get X cubed d excess in detail before now, let's look at our integrate inte grant. This caused X rays. How how well are indignant Look like in terms off t so cause X rays to the power forward has replaced by t and we have execute DX. And from this equation, execute DX is nothing but GTO before so 1/4 comes outside that outside that it is a constant integral off causes sign and we also have a constant off. Integral A c uh, where the T s X rays to the powerful. So we have one over for sign off X rays to the ball four plus C, which is actually the same former which Waas asked in the which was mentioned in the question as well. So the final answer is one or four times. Sign off X rays to the portfolio, plus C.

This is our number 36 in which we need to elevate. Excuse It is the products of square DX First by first we need to substitute something and then to then we need to use indication by parts. So let us rewrite is as access Squire and two x It is the product of Squire DX and let us multiply and divide by two. Okay, Now let us substitute it is to the power access choir equal to t. So it is to the products Squire in two different season of X square, that is to works the X equal to d. T. So this is to works. It is the Power access Squire, the X equal to DT. And one more thing from here. We had resumed areas with Pradaxa Square equal to t. So let us take Ellen both the sides. Yeah, Okay. Ellen De So this will become access square equal toe. Alan T now it's substitute teach and everything over here we will be getting Yeah, our interior equal to one by two. Ah, di l n t DT. Okay. Oh, not just simply TV t o simply, uh, this will value of this value This is only DT and value of excess square is simply Alan t. So this is Alan T DT. No, Let us assume this to be you this TV TV So we have you equal to guarantee Alan t. So do you will be equal toe one by t d t again Devi equal to DT So we will be equal to integration of d V and that integration of DT that is t not like it back Feel begetting Ah, one by two l n t DT equal to you into v according to when he gets on my part You into v So you is Alan T. Is he so t l anti mines video the and ah, do you? We have one y DDT. Okay. Into one by t d t. That is t l and T minus integration of d T s d and place constant medication K. Now that is, multiply both sides by two field plicating Ln t DT equal toe to T l N t minus one plus two. Okay, let us assume this to be C no, Alan DDT. Before substitution, it will become X cube. It is to the power access square D X equal toe to we had resumed D equals two. Okay, it is the products Esquire. It is to the products of square Alan T L N T means access choir from here. Excess square minus one plus c. So this will be the answer. Our Ah, but do no problem. This should be the answer. Thank you.

Neither. In this problem, we were asked to prove this identity. Uh, it looks nasty at first, but whenever we're asked to prove a certain anti derivative, the easiest way is always to start with the right hand side here, take its derivative and see if we end up with the into grand. So that's what we'll do. Let's take the derivative of Well, let's say the right hand side and see if we get back to where we started. So the derivative of this right hand side, the derivative of this first term, will have to use the product rule. So we'll get the first factor at times the derivative of the second factor, which is inverse, hyperbolic co Tangin X plus, uh, the derivative of first. And just to make it easy on ourselves, let's rewrite that X squared minus one all over to predators. Let's pull it apart. It'll just make the derivative easier times a second good. And then, of course, this derivative derivative of X over to Lord in Ella's 1/2 and a constant, of course, has contributed zero. Okay, ex scored minus one over to stays as it is from this chapter. The derivative of hyperbolic of inverse, hyperbolic co tangent. Let's just do the definition given in this chapter. And that derivative is one over one minus x squared for this derivative here and derivative of X squared over two is just x our room and everything else that is the same notice here. An X squared minus one one minus X squared weaken. We write this, Can you just pulling out a negative sign? The reason we're doing that is because things will simplify in the next. Step up everything else again. The same and just about at the end Here, Notice thes cancel. So we have minus 1/2 plus x times. Inverse, hyperbolic Oh, tangent X plus 1/2 but minus 1/2 plus 1/2 those cancel out. Final answer is X inverse, hyperbolic co tangent Becks. Also, since we started with the right hand side, we took the derivative and we ended up exactly with the into grand. We have proved that this integration formula is correct. Hopefully, this was helpful


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