Question

Hi, I need some help with the following integral.
I want to integrat between a and b the following function of x: Sqrt[ (x^2+1)/ x^2 ]

Answer 1

\[\large{\int\limits_{a}^{b}\frac{\sqrt{x^2+1}}{x^2}dx}\]is this ur question?

Answer 2

I will rewrite it: \[I rewrite \it; \int\limits_{a}^{b}\sqrt{(x^2+1)/x^2} dx\]

Answer 3

I would do a trig sub

Answer 4

Is the x^2 on the bottom in or outside of the square root?

Answer 5

\[\int\limits_{a}^{b}\sqrt{\frac{x^2+1}{x^2}} dx\]

Answer 6

Yes, it is inside!

Answer 7

Ok I will do go with a trig sub approach

Answer 8

still*

Answer 9

\[x=\tan(\theta)\]

Answer 10

Oh yes that works out nicely
Let me know if you get stuck anywhere

Answer 11

yes it does, doesn't it?

Answer 12

I dont get it ¨_¨ If I do that then I have to make a substitution and derivate Tg(theta) ?

Answer 13

\(x = \tan \theta \rightarrow dx = \sec \theta \tan \theta d \theta\) myin? /:) hahaha i think you mean sec^2 :P

Answer 14

Is getting more difficult at any step for me...Sorry but I am not able to see the path...it seems. = (

Answer 15

i'll let her do the talking...i've interrupted too many times already :S

Answer 16

@lgbasallote is right though, that dx should be sec^2

Answer 17

Other than that @Marrioncito just remember tan^2+1=sec^2 and then the problem will work itself all out.

Answer 18

now now let's let @myininaya do her talking now....she really hates interruptions...she might decide not to teach this poor kid if she gets mad...

Answer 19

yes sorry you are right

Answer 20

\[x=\tan(\theta) => dx=\sec^2(\theta) d \theta \]

Answer 21

int(sect/tant)sec^2tdt=int(1/sintcos^3t)dt=int(sint/cos^3t(1-cos^2t))dt
=int(du/u3(u^2-1)=...

Answer 22

x is so silly solution think about that i gave right

Answer 23

follow my solution you have to solve that!! Dx=1 !!

Answer 24

\[\int\limits_{\tan(a)}^{\tan(b)}\sqrt{\frac{\tan^2(\theta)+1}{\tan^2(\theta)}} \sec^2(\theta) d \theta \]

\[\int\limits_{\tan(a)}^{\tan(b)}\sqrt{\frac{\sec^2(\theta)}{\tan^2(\theta)}} \sec^2(\theta) d \theta \]

\[\int\limits_{\tan(a)}^{\tan(b)} \frac{\sec(\theta)}{\tan(\theta)} \sec^2(\theta) d \theta \]

\[\int\limits_{\tan(a)}^{\tan(b)}\frac{1}{\tan(\theta)} \sec(\theta) \sec^2(\theta) d \theta\]


\[\int\limits_{\tan(a)}^{\tan(b)}\frac{\cos(\theta)}{\sin(\theta)} \frac{1}{\cos(\theta)}\sec^2(\theta) d \theta\]


\[\int\limits_{\tan(a)}^{\tan(b)}\frac{1}{ \sin(\theta)}\sec^2(\theta) d \theta\]

\[\int\limits_{\tan(a)}^{\tan(b)} \csc(\theta) \sec^2(\theta) d \theta \]

Now I do integration by parts

\[\csc(\theta) \cdot \tan(\theta) |_{\tan(a)}^{\tan(b)} -\int\limits_{\tan(a)}^{\tan(b)}-\cot(\theta) \csc(\theta) \tan(\theta) d \theta \]

Answer 25

\[\csc(\theta) \tan(\theta)|_{\tan(a)}^{\tan(b)}+\int\limits_{\tan(a)}^{\tan(b)}\csc(\theta) d \theta \]

Answer 26

\[[\csc(\theta) \tan(\theta)]_{\tan(a)}^{\tan(b)}-[\ln|\csc(\theta)+\tan(\theta)|]_{\tan(a)}^{\tan(b)}\]

Answer 27

dont forget to sub back ^_~

Answer 28

I ve arrived to this integral as the final expression of a much larger exercise in what i had to calculate an line integral after giving a parametrization of a curve..demostrate that is regular, etc....is not fair that the difficult part relies in a one variable integral.
Thank you very much! =)

Answer 29

Because I hate using trigonometrics I have been thinking about and I have followed other path. Writing it \[\int\limits_{a}^{b} \sqrt{(x^2+1)/x^2}dx = \int\limits_{a}^{b} 1/x \sqrt{(x^2+1)}dx \] Then I made the subs:
\[u = \sqrt{x^2 +1}\] so \[\int\limits_{c}^{d} u^2/(u^2-1)du = \int\limits_{c}^{d} u^2/(u-1)(u+1)du\] And then integrate using partial fractions.
Is just another way...you have to do magic with that subs...but you don´t have to work with trigonomectric functions. :P