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Evaluate the integral.
$ \displaystyle \int_{\frac{\sqrt{2}}{3}}^{\frac{2}{3}} \frac{dx}{x^5 \sqrt{9x^2 - 1}} $
$\frac{8^{1}}{4}\left(\frac{\pi}{8}-1+\frac{7 \sqrt{3}}{16}\right)$
Calculus 2 / BC
Chapter 7
Techniques of Integration
Section 3
Trigonometric Substitution
Integration Techniques
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here we have a definite nickel off. One over X to the fifth times the square root of nine X squared minus one. Look inside the radical we see nine X squared minus one, which we can rewrite as three x and practices squared minus one. So that means is one. We should take three x to be one second data. That's our exam. So then X equals C can over three and then the ex is seeking data data over three potato. Also, we can go ahead and simplify this radical so nine X squared minus one is three x in parentheses, square minus one all inside the square root. And due to our tricks up here, we can write. This is seeking squared minus one and by the photographer, one of the protagonist identities This's tan sweater for the state of their and the square root of tents. Where is just change it also, This is a definite mineral. We see we have some limits here. There's our lower limit. So let's go ahead and find the new lower limit in terms of data. Now we have to observe that when we do a tricks up involving C can. The restriction for data is has to be between zero and pi resume or it's between pi In three two, we'LL see in a moment how this will help us. So let's plug in this lower limit for X. So using this equation up here so X is radical to over three for the lower limit. And this is supposed to equal the right hand side, which is she can't data, so this becomes seek and data equals radical, too. And that means that since data has to be in one of these two intervals, the only time this will happen is that they d equals power for and similarly, we could also find the new upper limit in terms of data. So to over three times three. So again, using this equation here. So there's our three x equals. He can't data. So we have See Cantero equals two. And the solution to this is Pirate Terry. So that's her no upper limit and new lower limit. It's uploading. These in tow are integral. We now have the integral from power for to pie over three d x. So we have seeking time, Stan, over three. It's over three and then debater. So that's the X and an excellent fit. So we go ahead and raise this expression to the fifth Power. So we'LL have seeking to the death. Well, the tree to the fifth and then we have nine X squared minus one in the radical. We've already simplified that expression down here in the bottom, right? And we had potential. So now we should just cancel out was much weekend. We see that we can drop one of the sea cans so that I'd become a four on the bottom and we could completely dropped its engines. And then we have looking at the threes. We have one over three over one over three to the fifth, and that's three to the fourth so we can go out and pull out a three to the fourth power and we're left over with one over seeking to the fourth power. So I'm running out of room here, so let me go to the next page. So three to the fourth is eighty one and a girl power for the laboratory. And then we can rewrite the instagram, this coastline of the fourth power. So here we can right Cose I into the fourth is co sign square and that's also square. And then we could use the half angle identity for coast and square. So the coastline square is equal to this fraction and we also square because this extra square on here and so we have one plus to co sign too. They don't plus co sign squared to theatre over four so was played by Listen, let's plot the four eighty one over four in a girl. Then we have one plus two co sign to data and then plus and then the very last term, we should also simplify use the half angle again. So for this term in the numerator, we could use the half Ingle again to right, This is one plus cosign for data over to you supplied in bed And over here we have one half co sign for data all over too Data. So I just plugged in this latest expression for co sign squared of two data and plugged it in. So here we could combine the one in the one half Add these together. So we have three house to co sign to data cosign for Daito over two and I already to integrate. And if this tooth aid and forth an hour bothering you and immigration feel free to use the use of here, you could take you to be too later here you can take you to before dinner. So after we integrate this, anyone over for three over to data, then we have sign of Su data plus sign for Daito over a and then our end points pie before Survivor three. So the next step will be to just go ahead and plug in these end points for they don't simplify it. Let's go to the next page to do that. So we plug in Piper three first. So three times data, which is five or three over two. Plus I know two theater. So that's two pirate three plus I knw of four pirate three over a. So that's plugging in power. Three. Now we plug in pie before and now we simplifies as much as we can. Here we know that sign a pious zero so we could drop this term so it's good and start adding things together. So we have eighty one over four. So here we see that we have a pie or two and in sign of to power. Three. It was really too. And in sign of four, pirate three, it's negative. Radical tree over, too. But then we all started in the bottom. So those are the first three. Then we have three papery plus one and then plus zero. So let's go ahead and add these fractions involving pie with Piper, too. And then we have minus three, five or eight. So putting these together, that's four pi over eight minus three pie operate. So we just have one pie. Brie. Such is from combining these two and watch out for the minus sign. Here we see that we have a minus one that doesn't cancel out with anything. So let's put the minus one. And then finally, we just go ahead and combined these radicals so get a common denominator, which is sixteen. So we have a radical three over sixteen minus one radical three over sixteen. So it's seven radical, three over sixteen, and there's a final answer
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