Integragtion Problem with Substitution

[Claire84]

At the moment I'm trying to solve this integration problem and it's not working out asd neatly as any other substituion problem I've tried before.

We have to integrate sqrtx divided by (1 + x^2) using the substitution x= tan^2t (as in tan squared t). So when we have dx/dt so you get 2tantsec^2t? I've tried using that but it all gets really muddled and complex. At the moment I'm down to finding the integral of 8 divided by -sin^2t but I'm not sue where to go from there. I've tried using substituions from this point on using the dpuble angle formulae but it's all going a bit pear-shaped. Any help would be much appreciated- thanks in advance!

 

[Muzza]

The integral of 1/sin^2(x) is -cot(x). Seems like the answer would be more complex...

 

[Claire84]

The answer we're supposed to get to is 2(sqrtx) - 2tan^-1(sqrtx)

 

[Muzza]

That's the integral of sqrt(x) / (x + 1) (notice the lack of a squared term in the denominator), which is "somewhat" easier than sqrt(x) / (x^2 + 1) ;)

We wish to integrate sqrt(x) / (1 + x) dx.

Let x = tan^2(t), which gives dx = 2tan(t)sec^2(t) dt.

The integral turns into sqrt(tan^2(t)) / (1 + tan^2(t)) * 2tan(t)sec^2(t) dt. The numerator is equal to sec^2(t) (standard identity). We can simplify it to:

tan(t)/sec^2(t) * 2tan(t)sec^2(t) dt =
tan(t)2tan(t) dt =
2tan^2(t) dt

Hopefully you can handle the rest.

 

[Claire84]

Really? Eeek what's wrong with this sheet then? It tells us to use the substitution x=tan^2(x) so would we not then need this for the one you've suggested? V.confusing.

 

[Muzza]

Added more info to the last post

Btw, try inputting Sqrt[x] / (x^2 + 1) at http://integrals.wolfram.com/ , and see what you get... Not the kind of thing you'd want work out for yourself on paper

 

[Claire84]

LOL, while you were typing that I was off tryingit myself and I've got down to the 2tan^2(t) bit! After that is there an integral you can use or do you have to change it to sin^2(x) over cos^2(x) and then use the double angle formulae? It just looks messy everytime I work with it? God, I should really remember to bring my brain home with me from uni!

 

[Muzza]

Personally, I'd rewrite it as tan^2(x) = (1 - cos^2(x)) / cos^2(x) = 1/cos^2(x) - 1 and work from there.

 

[matt grime]

Originally posted by Claire84
At the moment I'm trying to solve this integration problem and it's not working out asd neatly as any other substituion problem I've tried before.

We have to integrate sqrtx divided by (1 + x^2) using the substitution x= tan^2t (as in tan squared t). So when we have dx/dt so you get 2tantsec^2t? I've tried using that but it all gets really muddled and complex. At the moment I'm down to finding the integral of 8 divided by -sin^2t but I'm not sue where to go from there. I've tried using substituions from this point on using the dpuble angle formulae but it's all going a bit pear-shaped. Any help would be much appreciated- thanks in advance!

looks like a simple misprint, just use x=tan(t)

for questions like this, try and fiddle cos^2 + sin^2 =1 so you get something that looks like what you want

 

[Claire84]

Hm okay, so if we had 1 over cos^2x - 1 would you then have to change that to -sin^x? You gave the integral of this earlier, well the integral of 1 over sin^2x which was -cotx, but how did you work that out? I'm generally fine with differentiating but integration is really confusing me, esp. when you have 1 over a trigonometric function.

 

[Muzza]

Originally posted by Claire84
but how did you work that out?

Looked it up I'm not aware of any way to "construct" such trigonometric integrals, other than simply checking the derivatives of the standard trig. functions (i.e sec(x), sec^2(x), csc(x), etc) and seeing if something fits.

 

[Claire84]

Hm but how do we get the correct answer then if we have the integral of 1 over -sin^2x?

 

[Claire84]

Okay if we end up with the integral being -cot(t) then what do we do? How do we get x back into it from what we made x equal to?

 

[Claire84]

I end up with 1/x. Odd.

 

[matt grime]

Originally posted by matt grime
looks like a simple misprint, just use x=tan(t)

for questions like this, try and fiddle cos^2 + sin^2 =1 so you get something that looks like what you want

ignore me, didn't see the sqrt x in there

int 1/sin^2(t) is cot(t) give or take a minus sign.

x=tan^2t

so tan(t)=sqrt(x)

and tan is 1 over cot

so cot (t) is 1/sqrt(x)

 

[Claire84]

Okay, I just need to know how you get from the integral of 1 over -sin^t to 2sqrtx - 2tan^-1(sqrtx). It just doesn't seem to work.

 

[Muzza]

Looks like there's some confusion here. I said that tan^2(x) = 1/cos^2(x) - 1. The -1 is /not/ in the nominator. I meant (1/cos^2(x)) - 1, or sec^2(x) - 1. The integral of this is tan(x) - x (found a table of integrals here: http://www.math2.org/math/integrals/tableof.htm), so the integral of 2tan^2(t) dt (please excuse the sudden change of variables) is 2tan(t) - 2t.

Since we let x = tan^2(t), get that that sqrt(x) = tan(t), or t = atan(sqrt(x)).

Which gives 2tan(t) - 2t = 2tan(atan(sqrt(x))) - 2atan(sqrt(x)) = 2sqrt(x) - 2atan(sqrt(x)).

Plus a constant. ;)

 

[matt grime]

Originally posted by Muzza
Looks like there's some confusion here. I said that tan^2(x) = 1/cos^2(x) - 1. The -1 is /not/ in the nominator. I meant (1/cos^2(x)) - 1, or sec^2(x) - 1. The integral of this is tan(x) - x (found a table of integrals here: http://www.math2.org/math/integrals/tableof.htm), so the integral of 2tan^2(t) dt (please excuse the sudden change of variables) is 2tan(t) - 2t.

Since we let x = tan^2(t), get that that sqrt(x) = tan(t), or t = atan(sqrt(x)).

Which gives 2tan(t) - 2t = 2tan(atan(sqrt(x))) - 2atan(sqrt(x)) = 2sqrt(x) - 2atan(sqrt(x)).

Plus a constant. ;)

dunno if I'm helping or not, but, muzza, you've inconsistently muddled up t's and x's throughout these posts. perhaps claire starting again from the beginning would help along with her posting all her working out up to the point she gets stuck.

 

[Muzza]

Yeah.

 

[Claire84]

Okay, I'll go scan in my working out because it's easier to see on paper!

 

[Claire84]

Okay, couldn't get it small enough. Really not good with computers. Here's what I have.

I've got that dx=2tantsec^2(t)dt

so overall, the integral from before with the substituion x=tan^2(t) is
tanx divided by a+tan^2(t), and this whole thing is multiplied by the new value of dx with the dts in it that I mentioned above. I've got that all this cancels down to the form where we have the integral of 2tan^2(t), and then you could take the 2 outside of the integral sign but I'm not too sure what to do next.

 

[matt grime]

ok, we agree (up to multiplying by a minus sign at least,) that you want to integate tan squared

tan squared is sin^2/cos^2

ie

(1-cos^2)/cos^2 = -1 + sec^2


so integrating that wrt t (forget the factor of two out the front) gives us -t + cot(t)


recall tan^2(t) = x so tan(t) = sqrt(x)

t = tan^(-1)(sqrt(x))

cot(t) = 1/sqrt(x)


so the integral (again forgetting that two)


is -1/sqrt(x) + tan(-1)(sqrt(x))


put the two in (I think it's minus two cos diff tan gives -sec^2, not sec^2)

gets you close to what you want. not sure about the 1/sqrt(x) - can you check the answer sheet, does it say x to the half or x to the minus half?

 

[Claire84]

The exact answer that is given is 2sqrtx -2tan^-1(sqrtx). So both square roots are positive.

Oh and I never knew that integrating sec^2(t) gave you cot(t). Is that one of the integrals you're supposed to remember? Is it easy enough to work out?

 

[Claire84]

As in x to the half as opposed to the minus half.

 

[Claire84]

So there's definitely no way that the x on the bottom line could be x^2 then? Just so I can let people know at uni tomorrow cos I know a lot of them were getting a bit agitated at this question...

 

[matt grime]

well, it could be, but that would be a nasty integral to solve, seriously nasty, and it is so close to the one with x not x^2 on the bottom... mistakes happen on homework sheets.

 

[Claire84]

The lecturer today confirmed today that is was x on the bottom line as opposed to the x squared. But still don't have correct answer. Major brain frazzlement!

 

[Claire84]

Have got the right answer, I think! It's actually from one of the previous posts here but I've got that the integral of sec^x is tanx as opposed to cotx. Hope that's right. If not, don't spoil my excitment at actually having possibly got something right.