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The integral of a partial fraction expansion of the form A B Cr + D + I + 13 (2 + 13)2 22 + 36may result in a trigonometric substitution if which coefficient is not...

Question

The integral of a partial fraction expansion of the form A B Cr + D + I + 13 (2 + 13)2 22 + 36may result in a trigonometric substitution if which coefficient is not 0?There would not be a need for a trig substitution to integrate this expression;

The integral of a partial fraction expansion of the form A B Cr + D + I + 13 (2 + 13)2 22 + 36 may result in a trigonometric substitution if which coefficient is not 0? There would not be a need for a trig substitution to integrate this expression;



Answers

Integrate $\int \sec \theta d \theta$ by
$$
\begin{array}{l}{\text { a. multiplying by } \frac{\sec \theta+\tan \theta}{\sec \theta+\tan \theta} \text { and then using a us-substitution. }} \\ {\text { b. writing the integral as } \int \frac{1}{\cos \theta} d \theta \text { . Then multiply by } \frac{\cos \theta}{\cos \theta}} \\ {\quad \text { use a trigonometric identity and a u-substitution, and finally }} \\ {\text { integrate using partial fractions. }}\end{array}
$$

Uh this one here is gonna be hard. That's hard. But I just you know worked it out in in a paper and it is messy. I think it's going to take a lot of space here. Like the last december before this one No the last to determine before this one. So I'm just gonna cut things short and then uh you know we move ahead so uh you want to perform those integration right? You have to make a substitution because during trigonometry functions and I just I mean it's not great to work in trigonometry functions whenever you want to do uh simple algebraic stuff. So what you want to do is let you know you're gonna make some sense, you're going to let you be uh signed data. So perform a you saw let you be signed theater. What does he do you then do you? It's just gonna be co sign, they really did a key. So when you perform a U. Sub. Is gonna make uh you know things a little uh friendly for you to work with. Right? But whenever you make a use sub, you suppose this is a definite integral is not an indefinite integral is a definite integral. So whenever you make a use up you're supposed to change the limit of integration, right? So I'm gonna change the limit of integration. Whatever um theater is one of the theater is zero. When you put zero here, what did you use? Just gonna be zero? Right? Because sign 00 Whenever Theatre is Pi over four, What is you gonna be? Well, you is going to be signed prior before and that is one over root two. Right? So you is gonna be one over root two. One of our theater is power before. So the upper limit is going to be possible. The new upper limit is gonna be one or two. And then the new lower limit is still gonna be cereal. Okay? After the substitution. So what we have here is the integral of zero and 1 over uh Route two. And then the substitution is going to give us one over, you know, one minus U squared, one minus U squared. And then you uh squared to the school. Okay? And then this is gonna be do you? So this is a an integrated which is a rational function. Okay? And we're gonna treat it, we're gonna decompose it into partial fractions. Okay, This is just the same as the X. Would be using right? You can just view this one X where it's not it's not a problem. So this is gonna be one over now. This one can be actually it can be factored to be one minus you one plus you. Okay? Remember I said a squared minus B squared is a minus B. A plus B. Remember. So that is what I just did. But this one I mean you can't uh you can't do that to it because it's not minus is plus and just a square here. So that is going to stay. It's just gonna stay. It's gonna be a quadratic factor. So then this is gonna be a over one minus. You be over one plus you. Now this one is a quadratic factor and is repeated as well. Quadratic and repeated. So it's going to be see explosive T over uh U. Squared plus one and then E. X plus F. Over Use Great Close one. How about that? Mhm. Mhm. Soon we perform the partial fractions. You can have these us the constant. Right? So this entire thing is going to be I'm just gonna replace him. Right? So a is what a great B. S. one for 8? Mhm. C. is zero DS 1 before? since C0, I'm just gonna Wipe This one clean and then d. Is just one of the four. So I can bring this for here. E zero Fs one half. So I can do same. Right? We're just lucky that these are zero. So we have this thing here and I want to find the integral of that. Right? So you're just gonna integrate? Uh Right, it's a definite integral. I'm just gonna bring the limits of integrations later. Right, So we're integrating this thing here in a term by term Basis. Right? This one is 1/8 Natural Log of one. One minus you. Okay. And then the next one is 18 one or 8. Right? This one is -18 Natural Log of 1- You. The next one is Uh positive 1. 8. No, actually I love one plus you. Mhm. Mhm. Right, that is the second term. Third one is uh you know, 3rd 1 is one of the four inverse tangent of just you. Right. I told you this 11 over this is inverse tangent, right? So it's just bring out the one before as a constant term and they have inverse tangent of that. And then you have one of the four inverse tangent of you. So one of the four. That is this one right? One of the four uh inverse tangent of U plus U squared U over U squared plus one suit goobers Tangent of you plus you overuse squared plus one. So that is the integral of the entire thing. Right? If you want to do this integral uh you can have you have to do a trigonometry substitution. Where we did that in previous tutorial. Right? You do trigonometry except substitution. You're gonna let if you let oh uh you know if you let uh u be tangent t let you be equal to tangent theater. If you do this substitution here you can perform the youtube. Uh the integral. Right? We did that. I showed you how to do trigonometry integral in a very long tutorial. Right? So when you do that, do that substitution, you're going to get this one as the integral. Now this is the indefinite integral. But you want to find a definite integral. You wanna if I would this thing as a problem it and lower limit. Right? I'm a little pressed for space. I'm just going to use this one. Yeah, The low limit is zero and the upper limiters that, right? So whenever I see you I'm gonna put that's one their first and then minus where I see you, I'm gonna put zero. Right? So this is just straight out calculator work. So I'm just gonna put in the answer, right? Just put into your calculator and you're gonna approximately you're going to have Okay, 8.65 as your answer. Okay? So put this one in the calculator, You're just gonna put this one and we please you well, one over descriptive, too. And then minus then you put the same thing again. But this time you replace you with uh zero, right? And then you just uh performing on a calculator. And didn't you just have this one right as the answer to the question? Okay.

To integrate this, we don't need to use triggered a metric substitution because U substitution will work. And here we want to let U equal to X squared plus four X plus 13. And we get differential of U. Equal to to expose four. And then dx. So by substitution this is just equal to the integral of D. U. Over you which is equal to Ln absolute value of you plus C. Now since U S X squared plus four, expose 13 then this is equal to Ln absolute value of X squared plus four X Plus 13. And then pussy. And since X squared plus four express 13 is always positive. Then we can get rid of the absolute value and just right Ln of X squared plus four X plus 13 and then plus C. So this is our indefinite integral.

To integrate this. We will first apply completing the square. So we will reread this into the integral of one over the square of X squared plus six. X plus nine. And then plus four. And then the X. This is equal to the integral of one over the square of X plus three squared plus two squared. And then D. X. And then from here we apply triggered a metric substitution. We let X-plus three equal to tangent data so that the X. Is equal to two seconds squared. Terra di terra. So from here we have The integral of one over the square of two Tangent data. Square. That's for tangent squared data plus four. And then the X. Will be replaced by two seconds squared parody Thera. And then from here we have the integral of two seconds squared. Theta over. Yeah the square of four times tangent squared. Theta plus one. And then we have decided to. And since tangent squared Theta plus one is just seek and squared ferre. Then from here we have the integral of two seconds squared. Theta over. We have 16 times Seeking theories to the 4th power. And then the data. And we can cancel out the two from 16 and then you have eight years. So it also cancels out the second squared From seeking to the 4th sarah. And we're left with second squared theta. So this will be equal to 1/8 integral. Love want over seconds where tha tha that's just coastlines squared. Sarah de cera. Now you sing the half angle identity for ghosts and square data. Then you have 1/8 Integral of one plus Co sign of two. Theta over to and then decided to and this is equal to 1/8 Times 1/2 integral of one plus go sign of two theta and then decide to. So from here we have 1/16 times stayed A plus sign of to sarah Over two. And then plus C. And since sign of two theories to sine theta. Cosign theta. Then from here we have 1/16 Theta plus 1/16. Sign for a cozy and sarah. And then plus C. Now since say it is not the original variable. Then we go back to our substitution that this year X-plus three equals two tangent. Theta this means that tangent Theta yes Express 3/2. And theories just tangent inverse of Express 3/2. If we draw our right triangle with data here since tangent ferris opposite over the adjacent side. And this is express three. This is two. And the hypothesis will be square it of the square of X plus three plus the square of two. That's just X squared plus six. X plus 13. And so from here we have sign a theta equal to opposite over the hypothesis. So that's expose three over squared of X squared plus six X plus 13. And co science era will be equal to the adjacent side over the hypothesis which is squared of X squared plus six X plus 13. So from here we have this equal to 1/16 times tangent inverse of X was 3/2 Plus 1/16 times Science and just Express three over the square root of X squared plus six X plus 13 times Go sign theater was just too over square root of X squared plus Takes X was 13. And then pussy. What? You can simplify further into 1/16 tangent inverse of X-plus 3/2 plus we have X-plus three over eight times X squared plus six X plus 13 and then plus C. So this is our indefinite integral.

In discussion we need to find the value of integration X into arc tangent X dx with the help of integration by parts and by that agreement substitution. Let's see how to hold this question. Consider use the calls to are 10 X and D vehicles to X D X. Therefore by the replication we can right Do you with equals two Bx upon one plus access choir and by the integration of DV we can right were the calls to access square upon to Now we have the values of you we D U and D V. Therefore apply the formula of integration by parts which is given by integration ut vehicles to ub minus integration video So the Sylvia calls to integration X into are ancient texts. Do you have physical too? You want to be? That means X squared up on two. Ah tangent tax minus one upon to integration access square upon one plus x esquire V X. This expression can also be prettiness access square upon to are tangent tax minus half. Integration 1 -1 upon one plus x esquire D F. Now interviewed. Dysfunction. So the integration will be X is squared up on two. Uh tangent X minus x upon two plus 1.2 are tangent X plus a constant of integration. Hands the integration X are tangent X. D f will be close to one upon to Axis Square plus one are tangent tax minus X upon to plus a constant of integration. So this is the final answer for this problem. Now let's come to part B. Yeah now we need to find the integration of same function X. Are tangent. Tax dx with the help of programmatic substitution. So let's see out of all this consider X equals two. Then gently you therefore you will be close to are tangent X. And by the differentiation of X we can right. Do you accept the calls to? So you can't esquire you do you? Mhm. Hands the integration X. Arc tangent X. Dx can be written as integration. You then gently you he can't esquire you do you? No to solve this integration. Consider share difficult to you and DV recalls to tangent you he can't esquire you do you therefore by the differentiation of God we can right? Yeah. DJ are difficult to be you and by the integration we can write we're equals two. One upon to tangent square. You now I apply the integration by part so we can write integration and shared D. V. If he calls to Jared b minus integration we deserve. Mhm Hands the integration X. Are tangent tax. D. S will be close to you buy two tangent esquire U -1 integration tangent esquire you do you? So this will be calls to you buy two tangent. Esquire you -1 upon to integration tangent square. You can Britain us He can't esquire U -1. Do you now integrate this function so we can write you buy two tangent esquire U minus half tangent U minus U plus a constant of integration. So this will be calls to you buy two can gently Squire you minus Tangent. You buy two plus You buy two plus constant of integration from here. You buy two can be taken at the common. So this will be you by two into Tangent, Squire you plus one minus Tangent. You upon two plus constant of integration. Now substitute the value of you from relation act articles to tangent you. So the expression will be integration X. Arc tangent tax D X. If he calls to one upon to one plus x esquire are tangent tax minus X upon to plus constant of integration. So this is the final answer for this problem. I hope you understood the solution. Thank you.


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