integrate-each-function-by-using-the-table-in-appendix-d-int-6-cot-4-2-d-x-2

Question

Integrate each function by using the table in Appendix $D$.
$$\int 6 \cot ^{4} 2 d x$$

Amy Jiang · Accepted Answer

your case. They&#x27;re using her online computer. We get the following for Integral.

R M · Answer

problem. Number 40 Configuration off X A configuration of excess square plus six x over X squared for three. Power to be X, which is equal to integration off X squared plus three over X Squared the block three forward to the X plus integration off six x over X squared the for three for two, the X minus invigoration off three over X squared, both 3/2 and the ex Ah, this can be simplified into integration off one over X squared per three The x both three invigoration off two x over X squared for 3/2. The X minus three invigoration off one over X squared of both three. It&#x27;s great e x, and this is equal to so the first integration here can be sold with using poor Jamuna 32 from Beach T two. Um, so the answer for that is equal, and the second integration can be evaluated using a substitution as you is equal to X, it&#x27;s squared plus one. So you is equal to ah to Rex the X ah, so the integration can be written as integration or one one over you squared you and the third integration can be evaluated using Formula 33 on Page de to. So the answer for the integration for the three integrations is equal to one over the square. Root for three, then in verse over X. Over is growing with three miners, three over X squared, the for three minus X would two X square for three minus 1/2 square root for three, then in verse, X over square root or three process.

Amit Srivastava · Answer

seek Full X six is a constant sort. Take this outside and we can ride the second squared X into second x x So secret one place, then good x We&#x27;ll take great X the X So finally, this is cake 60 x and six tennis x take a square x 30 x If we go for it on the way we did it in Dickerson will be 10 x and here If we juked and e so this will be t square. The team can t to stake a teepee. So our debate answer will be extend x Plessix the three by three once a 6 10 x blessed to and cue it by three Bless.

Andrija Isakov · Answer

in this question will be finding the integral of X times two X quick two to the power of X quick. So first notice that we have an X squared was derivative is two X and we have X dx in the expression as well. So we noticed that we have a function inside of function and the inter function also has a derivative present. So that gives us the hint to use substitution. And we looks may substitute you for X squared, which means the value of the integral or the interval can be rewritten as to to the exponents you times 1/2 the year. This, of course, equals 1/2 of the integral of to to new to to the U, which by our integration table by formula equals to the exponents you over lawn of to plus the constant of integration. And so no, we have found the indefinite integral. We found, um, the family of anti derivatives. I would correspond to our original integral. Now we just have to substitute in terms of the variable. We had a beginning X since U equals x squared. You can write this us two to the power of X squared minus one all over long enough to close the constant of integration

Kelly Maluccio · Answer

So today we will be solving the integral from 2 to 6 of either the two x plus three y d X. Now the first thing we do when we look at this is we we can figure out howto integrate e to the X. Something simpler. So we want to use what&#x27;s called u substitution. And to do that, we have to pick what&#x27;s Are you gonna be So are you in this case is going to equal to X plus three. Why? And that means that our d&#x27;you has to equal, um, two times DX because three wise like a constant and so the derivative of that in terms of X will be zero. So now I have my you and might do so I can rewrite my integral here, and I&#x27;m going to write just a definite interval and then we&#x27;ll bring back the two in the six. So this will be the mineral of E to the u times D&#x27;You And don&#x27;t forget about this two over here, we have to make ah, 1/2 and I&#x27;m gonna bring this in the front. So since it&#x27;s a constant, we can take it out, and it could be 1/2 times this interval of E to the U D U. So now we know how to integrate this because the inter role of either the U is just eat of the U. Um, if you don&#x27;t remember this, it&#x27;s also the interval of or it&#x27;s e to the you and then over natural log of your the base. But the natural log of ease Just one. And so that&#x27;s why he is a special function where when we integrate this, we just get 1/2 e to the U. Normally, if we were just solving the definite and a role we would put plus C at the end. But we don&#x27;t need that here because we&#x27;re gonna end up plugging in this, too. In this six sow, the seed would cancel. So I&#x27;m gonna write 1/2 e to the you, which I&#x27;m gonna replace my you back with two X plus three. Why? So say two X plus three. Why? And I&#x27;m gonna solve this from 2 to 6 or integrate this rate from from 2 to 6. So first, what we&#x27;re gonna do is plug in the six for X. Okay, so we&#x27;re gonna get 1/2 times E to the to 10 6 which is 12 plus three. Why and then minus. And now I plug in the two. So I have 1/2 again times E to the two times two, which is four plus three wives. And now we can write this a little bit nicer just by writing the 1/2 once. And you could have done this to start. But now we&#x27;ll have 1/2 and then we&#x27;ll do e to the 12 plus three. Why minus e to the four plus 31 And this will be our answer for our indefinite integral above right here.

Problem

Integrate each function by using the table in App…

Problem 44 Easy Difficulty

Integrate each function by using the table in Appendix $D$.
$$\int 6 \cot ^{4} 2 d x$$

Answer

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Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus

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Basic Technical Mathematics with Calculus

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Allyn J. Washington, Richard S. EvansBasic Technical Mathematics with Calculus

Heather Z.

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Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus