Question

Still having trouble integrating a problem with trigonometric identities and a few instances of u-substitution. Posted below in a minute.

Answer 1

\[\int\limits_{}^{}\frac{ 1 }{ (2x-1)\sqrt{(2x-1)^{2}-4} }dx\]
\[u = (2x-1), du = 2dx\]\[\frac{ 1 }{ 2 }\int\limits_{}^{}\frac{ 1 }{ u \sqrt{u^{2}-4}}du\]

Answer 2

One minute, my computer's freaking out.

Answer 3

Posts are going to be piecemeal so I don't have to keep starting from scratch every time my browser crashes. \[u = 2\sec(x), du = 2\sec(x)\tan(x)dx\]

Answer 4

Alrighty, ill waitto see what else you come up with : )

Answer 5

Here's also where I start getting confused; just hopped on another library computer, hopefully this will be less buggy.

Answer 6

No worries.

Answer 7

I know that I'm supposed to make this substitution, but it's not like a "nomal" u-substitution wherein you take existing values and replace them with a single variable; here, we're taking a single, already u-substitute variable and replacing it else with something expanded, which is a little strange, but whatever. The way that I usually solve u-substitutions, I find an expression for dx in terms of du and du in terms of dx, x in terms of u and u in terms of x, that's where I start fumbling with this substitution:

Answer 8

\[u = 2\sec(r), du = 2\sec(r)\tan(r)dr.\]\[r = \sec^{-1}(\frac{ u }{ 2 }), dr = \frac{ du }{ \ 2sec(r)\tan(r) }\]

I literally keep trying to write an expression but I'm *REALLY* getting messed up about how to express this since I'm not substituting with a single variable, I'm substituting with a function and a variable.

Answer 9

Alright, lets see if this works out then. So we have u^2 - a^2, meaning u = (2x-1) = 2sec(theta) and the entire radical = 2tan(theta), which means we have this
\[\int\limits_{}^{ }\frac{ 1 }{ 2\sec \theta (2\tan \theta) }du\] and since u = 2sec(theta), du = 2sec(theta)tan(theta), which gives us:
\[\int\limits_{}^{}\frac{ 2\sec \theta \tan \theta }{ 4 \sec \theta \tan \theta }= \int\limits_{}^{ }\frac{ 1 }{ 2 }\]
\[\int\limits_{}^{}\frac{ 1 }{ 2 }= \frac{ 1 }{ 2 }\theta \]Now we just need to get theta back into terms of x and y. So we said that (2x-1) = 2sec(theta). SO ifwe solve for theta we get:
\[2x-1 = 2\sec \theta = \frac{ 2x-1 }{ 2 }= \sec \theta \implies \theta = arcsec(\frac{ 2x-1 }{ 2 })\] So now we just replace theta with the above to finally get
\[\frac{ 1 }{ 2 }arcsec(\frac{ 2x-1 }{ 2 }) + C\]

Answer 10

I got confused about how you said that first line, but for absolute clarify I'm gonna try to rewrite what you said in my terms to see if I understood that right:
\[u = 2\sec(\theta), \theta = (2x-1)?\]
Theta replaces u, and the old u (now theta) becomes (2x-1)?

Answer 11

*clarity

Answer 12

theta doesnt become anything until we solve for it in the end. What Im saying is that our u substitution equals TWO different things.
u = (2x-1) AND
u = 2sec(theta)

The point in saying this is because we need to realize that 2sec(theta) = (2x-1), which is important at the end.

Answer 13

Okay. So, yeah, on the same page. From there:, oh man, I don't want to say I'm scared of math, lol, but you're going to have to give me a minute because even though we're not, it feels like we're just throwing random stuff in here, and it's pretty much paralyzing me from moving forward, one minute

Answer 14

Alright, no problem. Obviously there is a systematic process to this, just gotta get used to it : )

Answer 15

\[u = 2\sec(x) = (2x-1); du = 2dx = 2\sec(x)\tan(x).\]\[\frac{ 1 }{ 2 } \int\limits_{}^{}\frac{ 1 }{ 2\sec(x)\sqrt{(4\sec ^{2}(x)-4} }dx\](would that be with respect to x???)

Answer 16

Respect to theta. We're using triangle identities to solve these problems, so we keep our substitutions in terms of theta. I mean, its not going to matter in the end as long as you dotn confuse yourself. But yeah, I see you do an extra step that I dont do in my work. Where'd the 1/2 come from ont he outside?

Answer 17

The original (1/2) came from the first u-substitution; du = 2dx, therefore dx = du/2 (expressing in terms of du, the denominator of the integrand gets a 1/2)

Answer 18

Nah, its going to confuse you if you start trying to do that. Yes, u = (2x-1), but differentiate it with respect to the trig substitution, 2sec(theta) and only do it with respect to the trig sub. So u = 2sec(theta) du = 2sec(theta)tan(theta). Dont worry about differentiating the (2x-1) part.

Answer 19

(I didn't just totally figure it out, but thanks for sticking around, lol, and yeah, I'm a total expert at confusing myself.)

Answer 20

Right, lol, no worries. So we find 3 things
u = _______
du (with respect to the trig sub) = _______
radical expression = ________

then wedo our substitutions and integrate.

Answer 21

(Just so I can clarify what I'm pulling this from)

Answer 22

I think W|A is potentially expressing things in a way that they seem to be much more complicated than they actually are.

Answer 23

Yeah, even alpha does something extra. So Ill do the problem again and show all my work.

\[\int\limits_{}^{}\frac{ 1 }{ (2x-1)\sqrt{(2x-1)^{2}-4} }dx\]
u = (2x-1) = 2sec(theta)
du = 2sec(theta)tan(theta)

now alpha does an extra step, but when we have a radical of u^2 - a^2, we can skip to the assumption that the whole radical = 2tan(theta) So that being said, we have our 2 substitutions:

\[\int\limits_{}^{}\frac{ 1 }{ 2\sec \theta(2\tan \theta) }*2\sec \theta \tan \theta = \int\limits_{}^{}\frac{ 2\sec \theta \tan \theta }{ 4\sec \theta \tan \theta }= \int\limits_{}^{}\frac{ 1 }{ 2 }=\frac{ 1 }{ 2 }\theta \]
Since alpha uses tan, Ill use that, too.
\[\sqrt{(2x-1)^{2}-4}= 2\tan \theta \implies \frac{ \sqrt{(2x-1)^{2} - 4} }{ 2 }= \tan \theta\] meaning
\[\theta = \arctan(\frac{\sqrt{(2x-1)^{2}-4}}{2})\] meaning we get:

\[\frac{ 1 }{ 2 }\arctan(\frac{ \sqrt{(2x-1)^{2}-4} }{ 2 }) + C\]Now clearly this is different than the answer alpha gives, but answer can often be very different because of the trig involved. When I differentiate wolfram's answer on my calculator, I end up with the 2x-1 term on top. But the process I amdoing is correct.

Answer 24

Thank you, haven't read it yet, but that's exactly what I was about to ask.

Answer 25

Yes, all integrals with trig or ln in them can have many various answers that are all correct, just by the nature of the identities and properties they have.

Answer 26

Okay, so you just skip the initial u-substitution entirely, as it's sort of a useless middleman that's only necessary if you need to see how the expression can be written in order for the second u-substition; i.e., if I was doing this problem, I could do the first u-sub just to see other options to express the integrand with u-sub, and, if expressed in this manner where both u's equal the same thing in a certain manner, you could entirely disregard the first u-sub.

Answer 27

(One sec, reading the rest)

Answer 28

Yeah, I suppose. But there are definitely things you can skip in the process. I graphed both solutions and they are the exact same except wolframs answer is shifted down by about 1/2. which keep in mind doesntmatter because of the plus C. We have an unknown up and down shift with plus C, so as long as both of our answers are the same graph other than an up and down shift, we're fine.

Answer 29

This makes *WAY* more sense now after reading through your doing of it, thanks so much!!!

Answer 30

Yeah, np ^_^

Answer 31

I get how you did it, and now I'm just figuring out how to get to the book's answer (almost certainly through identities, I'm just now looking at it so I haven't necessarily not figured something out).

Answer 32

What is the books answer?

Answer 33

Bookanswer:
\[\frac{ 1 }{ 4 }\sec^{-1}\left| \frac{ 2x-1 }{ 2} \right|\]

Answer 34

Brb,one minute.

Answer 35

Back.

Answer 36

(I'm just looking for what kind of identity I should use; that answer's pretty friggin' different, and it's worrying that we're all coming to slightly different but nonetheless valid answers. I feel like if lazy instructors took a look and didn't immediately, see that the antiderivative was equal to the "right" answer, I would get downgraded. Not remotely your fault or problem, just an issue with standardized tests and math.)

Answer 37

Yeah, im guessing we do need to do the differentiation of (2x-1) like you were suggesting. I must just be forgetting something from forever ago, lol. Alright, let's do that. so u = (2x-1), so du = 2dx and dx = du/2
and since u also equals 2sec(theta), du = 2sec(theta)tan(theta)
\[\int\limits_{}^{}\frac{ 1 }{ 2\sec \theta (2 \tan \theta) }*\frac{ 2\sec \theta \tan \theta }{ 2 } = \int\limits_{}^{}\frac{ 1 }{ 4 } = \frac{ 1 }{ 4 }\theta\]
and since we said (2x-1) = 2sec(theta), we can say sec(theta) = (2x-1)/2 and:
\[\theta = arcsec \left| \frac{ 2x-1 }{ 2 } \right| \]giving full answer of:

\[\frac{ 1 }{ 4 }arcsec \left| \frac{ 2x-1 }{ 2 } \right|+ C\]so yeah, you did have to worry about differentiation u with respect to x and with respect to theta. Sorry bout that, mustve forgotten after s o long x_x

Answer 38

Oh, okay. Jeez, this was...a complicated problem, lol.

Answer 39

For some reason whenever I try to look at your profile, it doesn't show up -_-

Answer 40

Oh O.o Weird, lol. But yeah, Ill do another example to show ya if you want.

Answer 41

I'm good, I got this problem (But what are you talking about, do you have a suggestion of an example of something you want to show me)

Answer 42

Umm.....well, if you think you understand how to do it then thats awesome. But the example I was going to show was this:

\[\int\limits_{}^{}\frac{ x }{ \sqrt{x^{2} + 6x + 12} }dx\]

Answer 43

(Sorry, didn't mean to sound cocky, but I meant that I know that I understand this, at least to the degree that I relatively should; I don't know what I don't know and thus wouldn't know if there would be an example worth looking at.)

I'm gonna take a crack at that example for a minute and see where I get, if I can get anywhere.

Answer 44

Alright, sure. I think its a good example of an extra twist in these problems that should be known.

Answer 45

I'm guessing I have to complete the square, but I'm just extraordinarily bad at basic algebra, xDDD, gimme a sec.

Answer 46

Yes, complete the square, haha.

Answer 47

\[\int\limits_{}^{}\frac{ x }{ \sqrt{x^{2}+6x+12} }dx = \int\limits_{}^{}\frac{ x }{ \sqrt{(x+3)^{2}+3} }\]And then from here, I'm guessing u-substitution or something

Answer 48

(Btw, you'll never see me do integration by parts, I don't know why but I absolutely hate it, I mean, seriously, it's pretty awful/un-elegant IMO)

Answer 49

And if you have to do by parts? xD Lol, yeah, I like by parts, I think its easy to do x_x Its just yeah, there are situtations where you have to do it and it gets long.

Answer 50

I'm just super terrible at deciding v and du, etc etc. Okay, one sec.

Answer 51

alrighty.

Answer 52

\[\int\limits_{}^{}\frac{ x }{ \sqrt{(x+3)^{2} +3} }dx; u = (x+3),du=dx\]\[\int\limits_{}^{}\frac{ u -3 }{ \sqrt{u^{2}+3} }du\]

Answer 53

Okay, this might take me a sec, I have no idea what to do, lol..

Answer 54

I'm thinking of completing the square again.

Answer 55

Yeah, exactly why I bring this one up xD And nah, dont need to.

Answer 56

Let me do it just to see where it takes me, so I'll see where this goes wrong or becomes unnecessarily difficult for future reference.
\[r = (u + \sqrt{3}), dr = du\]...nvm I can see where the numerator will get messy.

Answer 57

yeah, this problem will get a slight bit messy for sure.

Answer 58

I'm guessing a trig identity now getting the denominator somehow to something like sqrt(x^2 + 1) or something. One sec.

Answer 59

Yeah, this is where skipping a step in these problems can be useful xD

Answer 60

\[\int\limits\limits_{}^{}\frac{ u-3 }{ \sqrt{u^{2}+3} }du = \int\limits_{}^{}\frac{ u-3 }{ \sqrt{\frac{ u^{2} }{ 3 }}+1 }\]

Answer 61

Uhhhh whoops.

Answer 62

So whats a?

Answer 63

(radical supposed to envelope the whole denominator, one sec.)

Answer 64

\[\int\limits_{}^{}\frac{ u-3 }{\sqrt{\frac{ u ^{2} }{ 3 }+1} }; s = \frac{ u }{ \sqrt{3} }, ds = \frac{ 1 }{ \sqrt{3} }du\]\[\int\limits_{}^{}\frac{ \sqrt{3}(s \sqrt{3}-3) }{ \sqrt{s^{2}+1} }ds\]Somewhere I messed up that last expression regarding a constant multiple from the whole du=ds business, I regularly do that.

Answer 65

You just dont like using the trig subs, lol.

Answer 66

I HATE USING THEM lol, but is that right so far? I actually don't think I made a mistake on that, retrospectively.

Answer 67

\[\int\limits_{}^{}\frac{ 3s-3 \sqrt{3} }{ \sqrt { s^{2}+1} }ds = ...\]

Answer 68

But hey, I'm about to use a trig sub if I fiddle with this last part, right? One sec.

Answer 69

Its pretty weird for me to see it with all the u and s subs like that, haha. Not sure why you did a double sub really O.o

Answer 70

\[\int\limits_{}^{}\frac{3s-3 \sqrt{3}}{\sqrt{s^{2}+1}}ds = \int\limits_{}^{} \frac{3s}{ \sqrt{s^{2}+1}}ds-\int\limits_{}^{}\frac{ 3 \sqrt{3} }{ \sqrt{s^{2}+1} }ds\]

Answer 71

\[\int\limits_{}^{} \frac{3s}{\sqrt{s^{2}+1}}ds = 3\int\limits_{}^{}\frac{ s }{ \sqrt{s^{2}+1} }\]\[\int\limits_{}^{}\frac{ 3 \sqrt{3} }{ \sqrt{s^{2}+1} } = 3 \sqrt{3} \int\limits_{}^{}\frac{1}{\sqrt{s^{2}+1}}\]

Answer 72

Im not sure quite where this is going, so I gotta see your trig sub to know if its wrong xD

Answer 73

Uhh, now the top one seems to be weird, what can I do with that, but the second one is straightforward, amirite?

Answer 74

I just know you dont really need a double sub as far as I can tell.

Answer 75

LOL through the bad jungle of unncessary u-substitution

Answer 76

Pretty much o.o You started off doing u, then went to r, then ditched r and went to s, lol.

Answer 77

Wow, guess I totally forgot that there wasn't a trig identity with sqrt(x^2+1) in the denominator, or is there? I really feel like there's an identity for this but I can't find one.

Answer 78

If you have sqrt(x^2 + 1), its going to become arctan for sure, but the answer has no inverse trig functions in it.

Answer 79

There's also arcsinh(x); d/dx arcsinh(x) = 1/(sqrt(1+x^2))

Answer 80

http://math.info/Calculus/Derivatives_Hyp_InvHyp/

Answer 81

Haha, yeah, you dont need hyperbolics for any of calc 2, haha.

Answer 82

I used them in calc 1!Not rigorously, but I used them, and I need to know them, along with the inverse hyperbolics, too, unfortunately. I'd rather not know them, honestly, lol.

Answer 83

(At least at the moment. It seems like unnecessary complication at the moment, especially since it's pretty much freaking impossible to my knowledge to evaluate some of the inverse hyperbolic functions without introducing complex numbers.)

Answer 84

But I want somebody to show me that what I've done so far is wrong; I can't see anything incorrect with my workings up until this point (although there might be), and it looks like I'd be able to end up with an inverse hyperbolic tangent. I'm gonna bump this, and in the meantime, we can take it the simpler way you wanted to (the way that actually makes sense).

Answer 85

Note: Although I have absolutely *no idea* what I'm going to do with that s/(sqrt s^2 +1) term off the top of my head.

Answer 86

Oh, darn! This isn't an open question. Oh well, let's just do whatever you were talking about.

Answer 87

So you want me to do the trig subs?

Answer 88

Yeah, lol

Answer 89

Okay, haha. Well you did the complete the square fine

\[\int\limits_{}^{}\frac{ x }{ \sqrt{(x+3)^{2} + 3} }dx\]Now here's the part we can skip. When I have the form sqrt(u^2 + a^2), I can automatically rewrite the whole radical as asec(theta)
u = (x+3)
x = u - 3
du = dx

also
\[u = \sqrt{3}\tan \theta\]
\[du = \sqrt{3}\sec^{2} \theta\]
\[\sqrt{(x+3)^{2} + 3} = \sqrt{3}\sec \theta\]
\[\int\limits_{}^{}\frac{ \sqrt{3} \tan \theta - 3 }{ \sqrt{3} \sec \theta }*\sqrt{3}\sec^{2} \theta = \int\limits_{}^{}(\sqrt{3} \tan \theta-3)(\sec \theta)d \theta\]

Can you see what ive done so far?

Answer 90

Whoah, sorry, got *totally* distracted for a second.

Answer 91

What's the equation for that identity again? Oh, and these are problems that were taught before we learned about that type of solving through those sqrt(x^2 + a^2) = asin/cos/whatever(theta).

Answer 92

(e.g. trying to see if can be solved without invoking that.) I'll show what techniques of integration have and haven't been covered, one sec.

Answer 93

In this chapter (first on integration in the book), The Fundamental Theorem of Calculus was covered, U-substitution, and...that looks like it, that was all thatwas covered so far, so that's what would be hypothetically available to solve this problem.

Answer 94

Alrighty. And here's the general list of trig subs:

when
\[\sqrt{u^{2} + a^{2}}\]
\[u = atan \theta\]
\[\sqrt{u^{2} + a^{2}}= asec \theta\]

when
\[\sqrt{a^{2} - u^{2}}\]
\[u = asin \theta\]
\[\sqrt{a^{2} - u^{2}} = acos \theta\]

and when
\[\sqrt{u^{2} - a^{2}} \]
\[u = asec \theta\]
\[\sqrt{u^{2} - a^{2}} = a \tan \theta\]

and you wouldnt know how to do this problem without some knowledge of inverse functions. This is usually taught before trig sub, but you have to know the forms of the inverse functions. Not hyperbolic, though, lol. I never did hyperbolic.

Answer 95

That's weird. This is in the chapter review, and yeah, I'm a little confused because in this chapter it looks like almost nothing allowing you to solve this has been covered, but it's there. Hm.

Answer 96

The only thing allowing you to solve this would have to be knowledge of what the inverse trig integrals look like. Before trig subs, they make you manipulate the equation to try and mimic the final form. I think its trickier than just doingtrig subs.

Answer 97

Yeah. One sec.