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QuestionEvaluate VI+sin?(1- 3x) sin(l - 3x)cos(l - 3x)dx by using substitutionQuestionEvaluate xInxdx by using integration by parts....

Question

QuestionEvaluate VI+sin?(1- 3x) sin(l - 3x)cos(l - 3x)dx by using substitutionQuestionEvaluate xInxdx by using integration by parts.

Question Evaluate VI+sin?(1- 3x) sin(l - 3x)cos(l - 3x)dx by using substitution Question Evaluate xInxdx by using integration by parts.



Answers

evaluate using Integration by Parts$$
\int 3^{x} \cos x d x
$$

Let's use a partial fraction the composition for this one. So first, before we do that, we should check if we can factor the denominator. And in this case, you can pull out of X left over with X squared plus three. Now, I would also have to check it this quadratic here in Prentice's if that matters, so you can try to factor this good. This one doesn't factor. And we get this by looking at the discriminative B squared minus four a. C. And this problem. There's no ex term in the quadratic. So be zero and then minus four times the coefficient in front of X squared is one. And then times three, which equals C. And this is a negative number. So that means that this quadratic will not factor. So using the case won for the linear factor. This's what the author calls case one non repeatedly near factor. And then here This is what the other calls case three. Because it's irreducible, quadratic. So in the numerator we need be explosive this time. And then that denominator X squared plus three. So now it's more supplying both sides of this new equation By that denominator on the left. So on the left side we have two ex ministry, but on the right, we have that. And then we can go ahead and simplify this. So let's pull out of X squared and we could pull a plus B and then we have CX and then we have three, eh? So I just combined these terms by factoring out some power of X. So here, x square, Then extend the constant term. And the reason for doing this is that we look at the left hand side. We see that there's no X squared over here, So zero x square. So on the right hand side and next to the X square, that's her must also be zero. So a plus B equals zero. Similarly, we have C equals two and then we have three A equals minus three. So this is two over here and then minus three over here. And so go ahead and saw these three. So here we have a CZ minus one. Then plug this into this and then you get B is one. And here we're from this equation we no see is too. So let's go ahead and plug this A, B and C, and so are partial. Fracture the composition. And then instead of integrating the left hand side, the original problem book integrate this right hand side. So let me go to the next page to write this. So a was negative one so minus one over X and then be was one. So one X and then see was too. So plus two, that's our inaugural. One way to proceed is to just right this into split this into three intervals. So I'm getting a little sloppy. Here, let me take a step back. Negative one over X. That's the second one. And then for the last one, we could even pull out that too. All right, so three in a girl's here, the first one that's just negative. Natural log, absolute value of X. For the second and roll. We can do a use of let you be that denominator. Then, after that, if you do to you and then divide by two, you get extra e x, which is the numerator. So this I have some use over here. So one half and then let me just write this in terms of X after you do the use of and right and back substitute. That's what you have. And then finally, for this last inaugural you saw was not going to work because of this extra X that we get the numerator. So for this one, we can do it tricks up X equals square root three tan data, then DX route three c can square data. So after doing this, tricks up for the third and rule and then rewriting everything back in terms of X, we have two, three over three and then we would have Seita. But then instead of so here, we would have a date after we integrate. But then we have ten data people's eggs over root tree. This is for martyring sub and then you could solve for data here by taking the are ten. So you would replace data and this expression here with our ten. And so let's write that and let me also finally let me add that seeing since we did integrate and then one half and actually on the second log, you could even drop absolute value since X squared plus three is always positive. And then to room three over three. And now we replace state of with this and boom, add that constancy and that's our final answer

Okay, so for this integral we're gonna let you We're doing integration by substitution. We're gonna let U equal three X -4. So then D. U. Is equal to three times dx. So that we can say also DX is equal to D you divided by three. And if U. Is equal to three x minus four then we can say X. Is equal to U plus four divided by three. And so we're gonna plug in um these values in for X and the numerator and then for three x minus four in the denominator here and infirm dx as well. And so this is equal to the integral of You plus four divided by three plus three divided by you um to the three halfs power. And then we also dx here is equal to D. You divided by three. So we need to um Multiplied by 1 3rd here and I'll just put that outside of the derivative. And so this is equal to um 1/3 times the integral. Have you divided by three plus four divided by three Plus three divided by you to the three of us power. D. You. And what we want to do is split this up into three different integral. Um Of the three different fractions that we're gonna have. So this is equal to one third times in parentheses the integral of four divided by and I'll actually take the four. Well we can do that afterwards four times 3 um times U. 23 halves power. And then oh sorry that's it that one's a you not a four. So you and then the second one is a four divided by three times you 23 halves to you. And then plus three divided by You do the 3/2ves do you? And so what we can do I guess first as we can minus the the power in the numerator from that in the denominator. Since our numerator has a lower power. So this is now going to become instead of three halves we're gonna have one half and this year is going to go away. And then the other thing we can do is just take out all the constants out of these integral. And so this is gonna be equal to one third times the integral of one third at times one third times the integral of it's gonna be U. To the negative one half power do you? And then plus four thirds times the integral of U. To the negative three halves And then plus three times the integral. A view to the -3/2ves do you to you? Okay so each of these integral czar. Just an anti r a power rule. Anti differentiation um integral. And so we add one to the power and divide by the power. That's all I'm doing here to find these integral. So I'm gonna have one third in parentheses one third times U. To the one off power. We add we add one to the power and multiply by the reciprocal or divided by that power. So that's multiplying by two. And then for the second one we have four thirds multiplied by U. To the negative one half power. And then we need to multiply by negative to here. So this is actually going to be eight and I'm gonna change this from a positive side into a negative sign. And then lastly for this last integral we're going to have plus three and then we add one to the power again. So we have you to the negative one half and then we divide by that power. Multiply by the reciprocal. So this one is also going to be negative and it's gonna be negative six times U. To the negative one half power. And then you can put this in parentheses and we can add our constant of integration plus C. However we do need to actually plug in what U. Is equal to. So sorry about that. I got a little bit ahead of myself so I'm gonna have one third times two thirds what was U. Is equal to use equal to three X minus four. So we have two thirds times three X minus four. It's the one half power -8/3 times um three x minus four. It's a negative one half and then minus six times three X minus four. To the negative one half. And then plus C. So this is the value of our integral.

Section eight down for a number of fifties, or we're dealing with a definite interval here that involves tricks substitution. So let's just rewrite this in a role. We need to do a little bit of algebra and brave manipulation here before we get started. So let's just write this 36 minus nine x squared to the three halves TX. And so what I need is the X squared needs a coefficient of one, so I need to factor what's in front of that so I can write. This is one over factor out in nine, and you left with four minus X squared. And then all of that raised to the three halves DX have been nine to the three halves that's going to be three cubed, which is 27. So this just turns into a 1/27. The integral 1/4 minus X squared to the three has DX. So now the substitution we're looking for in this case will be X equal to sign of data. Then D X will be too co sign of that at death, Anna and four minus X squared to the three halves that's going to be So that's four minus for sine squared of theta to the three halves. So that's 41 minus sine squared is coastline squared. So this becomes four to the three halves and then co sine squared of theta to the three halves. We're just going to be eight co sign cubed of Thalia. So this integral transforms into 1 27th And then what you see here it won over eight co sign cubed of theta, and the X is just too co sign of fate a d theta. So this turns into 1/1 08 one over co sine squared of fada D data, which is 1/1 08 seeking squared of theta. And we know that that just turns into 1/1 08 tangent of theta, plus a constant of integration. So our original substitution was X equal to sign data so x equal to scientific data. That means the sign of data is equal to X over two that makes this side four minus X squared square root. So this answer turns into 1/1 of eight. The tangent of the angle theta is just X over the square root of four minus x squared, plus a constant of integration

All right, so x times three to the X DX to integrate by parts. I will take you here and D V So you is X and do you is one TV is three to the X V will be three to the X over Helen off three. So moving on from there, we know that integral off you Devi is equal to u V minus into grow off v do you? So you ve that's X times three to the X over that line of three minus into go off three to the X over Ellen of three times one, which is do you so moving on from there? And that's x three to the X over Atlanta for three minus one over Elland of three times Integral off three to the X. As we know here, the integral If this is, um three to the X over Ln of three. So we have X times three to the X over island three minus one over. Ln three times, three to the X over Ln three. So expanding this x three to the X minus three to the X over Helen three squared plus constancy


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