How Do You Integrate the Square Root of Tangent?

[shaiqbashir]

PLease help me in solving the following integral!


[tex]\int{\sqrt{\tanx}dx}[\tex]


please help me as soon as possible because i want to submit my assignment tomorrow

 

[whozum]

$$\int{\sqrt{\tan x}dx}$$

I would imagine if you played around with it enough you could get it by parts. I can't see anything obvious though.

 

[Oggy]

Or
$$\tan{x}=t \Rightarrow \frac{dx}{(\cos{x})^{2}}=dt$$, and remember $$\frac{1}{(\cos{x})^{2}}=1+(\tan{x})^{2}.$$

 

[whozum]

$$x = arctan(u)$$

$$dx = \frac{du}{1+u^2}$$

$$\int \sqrt{ \tan x} \ dx \Rightarrow \int \frac{\sqrt{ u }}{1+u^2} \ du$$

Can you explain the discrepancy? I cant.

http://www.public.asu.edu/~hyousif/ma.JPG

 

[dextercioby]

There's no discrepancy whasoever.The 2 integrals are different.If you were to invert the sub.which you started with,the 2 results would coincide.

Daniel.

 

[whozum]

Oh, ofcourse. How could I forget that.

 

[PhysicsinCalifornia]

PLease help me in solving the following integral!


$$\int{\sqrt{\tanx}dx}$$


please help me as soon as possible because i want to submit my assignment tomorrow

Since this is an indeterminate integral, you have to find the anti derivative of the function $$f(x) = \sqrt{\tanx}$$

I would do what whozum did, which I like to call "u-substitution"

$$u = \tanx$$, so $$du = sec^2x dx$$ and $$dx = \frac{du}{sec^2x}$$

Is this right Whozum?

Using what I got, $$dx = \cos^2x du$$
Can u work from here ?

 

[whozum]

Yeah, you've derived the same result as I have, from there I think parts is the way to go.

 

[shaiqbashir]

Please help me a little more

Thanks a lot for ur precious help!

friends! I am still having a little problem in solving it. please can u people help me a little further to solve this integral.


thanks in advance!

 

[Jameson]

I can't provide any real help to the answer of this integral, but I did do it online and I've attached the answer. It's long, but maybe it will provide some insight.

Jameson

 

[Oggy]

Well it's not that difficult or ugly... After the substitution i pointed out it is simply:
$$\int{\sqrt{t}(t^{2}+1)dt}=...=\frac{2}{7}(\tan{x})^{\frac{7}{2}}+\frac{2}{3}(\tan{x})^{\frac{3}{2}}+C$$

 

[whozum]

Oggy the final integral is

$$\int \frac{\sqrt{ u }}{1+u^2} \ du$$

 

[whozum]

$$\int \frac{\sqrt{ u }}{1+u^2} \ du$$

$$g = \sqrt{u} \ \ \ h = arctan(u)$$

$$dg = \frac{1}{2\sqrt{u}}\ du \ \ \ dh = \frac{1}{1+u^2} du$$

$$\int g dh = gh - \int h dg$$

$$\int \frac{\sqrt u}{1+u^2} du = \sqrt{u}\ arctan u - \int \frac{arctan(u)}{2\sqrt{u}} du$$

 

[Oggy]

Yep u^2+1 should be in the denominator. Sorry

 

[shaiqbashir]

Thanks once again but what is its final answer

Hi Guys

thank you very much for ur tremendous help. You people are really very helpful.

Now whozum!

i have tried to solve this integral, from the point where u have left it. but I am getting a zero answer.

why is its so.

should i try to solve it further from the point where u have left it or should i stop it there. if further solution is required then what will be the right answer so that i should get it.

Please PLease help me! this question has now become a challenge for me.

Thanks in advance!

 

[dextercioby]

Here's how it's done.

$$I=\int\sqrt{\tan x} \ dx$$ (1)

Assuming the tangent is defined on an interval on which is positive,so we don't run into complex valued functions,one makes fisrt sub

$$\tan x=t^{2}$$ (2)


$$dx=\frac{2t}{1+t^{4}} dt$$ (3)

Then

$$I=2\int \frac{t^{2}}{1+t^{4}} \ dt$$ (4)

Use the trick

$$1+t^{4}=\left(1+t^{2}\right)^{2}-\left(\sqrt{2}t\right)^{2}$$ (5)

,so that (4) becomes

$$I=2\int \frac{t^{2}}{\left(1+\sqrt{2}t+t^{2}\right)\left(1-\sqrt{2}t+t^{2}\right)} \ dx$$ (6)

Use partial fractions and maybe partial integrations...

Daniel.

 

[GCT]

$$I=2\int \frac{t^{2}}{\left(1+\sqrt{2}t+t^{2}\right)\left(1 -\sqrt{2}t+t^{2}\right)} \ dx$$

Can someone check through Maple whether the answer to this is consistent with the original integral?

 

[dextercioby]

If you're doubting the calculations exposed above and,incidentally,not shortcut,why won't you put a pencil in your precious hand and grab a piece of paper and do it...?

Sides,Maple is a known f***-up.

Daniel.

 

[calculus1967]

I can't provide any real help to the answer of this integral, but I did do it online and I've attached the answer. It's long, but maybe it will provide some insight.Jameson

I don't think that's right; I think it should be:
$$\frac{-\sqrt{2}(\ln(\vert(\sqrt{2\tan (x)}+1)\cos (x) + \sin (x)\vert)-\ln(\vert(\sqrt{2\tan (x)}-1)\cos (x) - \sin(x)\vert)-2(\tan^{-1}(\sqrt{2\tan (x)}+1)+\tan^{-1}(\sqrt{2\tan(x)}-1)))}{4}$$

 

[calculus1967]

I don't think
$$I=2\int \frac{t^{2}}{\left(1+\sqrt{2}t+t^{2}\right)\left(1 -\sqrt{2}t+t^{2}\right)} \ dx$$
is consistent with the original integral, because that would be $-\cos^2 x$

 

[Jameson]

Well, that's what Wolfram's online integrator spat out. I couldn't defend it or not. It looks to be close to where this thread is leading with the partial fractions. Can you show the steps you used to arrive at your answer?

 

[dextercioby]

What "-cos^2 x"...?

Daniel.

 

[Jameson]

Actually now I'm fairly sure my answer is correct. Using Daniel's substitution and arriving at this...

$$2\int \frac{t^{2}}{1+t^{4}}\dt$$

I've attached the answer to this integral, and when you substitute

$$t = \sqrt{\tan x}}$$

back into this integral, you yield my original answer.

 

[calculus1967]

"-cos^2 x" is $-((\cos (x))^2)$

 

[Jameson]

I think Daniel was asking where did that come from?

 

[calculus1967]

Hmmmm... your probably right.

 

[Jameson]

So where did it come from? How did you arrive at your answer?

 

[calculus1967]

Which answer?

 

[calculus1967]

I am quite sure that
$$I=2\int \frac{t^{2}}{\left(1+\sqrt{2}t+t^{2}\right)\left(1 -\sqrt{2}t+t^{2}\right)} \ dx$$
is $$-\cos^2 x$$

I am much less positive about the other answer.

 

[calculus1967]

It's posible that
$$\frac{-\sqrt{2}(\ln(\vert(\sqrt{2\tan (x)}+1)\cos (x) + \sin (x)\vert)-\ln(\vert(\sqrt{2\tan (x)}-1)\cos (x) - \sin(x)\vert)-2(\tan^{-1}(\sqrt{2\tan (x)}+1)+\tan^{-1}(\sqrt{2\tan(x)}-1)))}{4}$$
is equivelent to your answer.

 

[Jameson]

Yes, it looks like it could be equilvalent.

 

[GCT]

If you're doubting the calculations exposed above and,incidentally,not shortcut,why won't you put a pencil in your precious hand and grab a piece of paper and do it...?

Sides,Maple is a known f***-up.

Actually I was working on this problem myself and thought it could be done by partial fractions after arriving at

$$2\int \frac{t^{2}}{1+t^{4}}\dt$$, I wasn't aware of the trick to progress towards the next step though

Your final form seems fine, however, I'll need to do some research on what makes such a partial fraction capable of integration (just finished int. calc this spring). Perhaps when I find the time, I'll try solving it p&p, just thought it could be done through Maple or Matlab. Isn't this a table integral (the OP's integral)?

It would also be appropriate for the final answer to be posted conclusively, if your final form proves to be right, good for you...and all of us, if not (I would be surprised if it isn't), than let's move on to see exactly why.

 

[dextercioby]

Yes.It's [1]:

$$\int \frac{x^{2}}{A+Bx^{4}} \ dx=\left\{\begin{array}{c} \frac{1}{4B\alpha\sqrt{2}}\left(\ln\frac{x^{2}-\alpha x\sqrt{2}+\alpha^{2}}{x^{2}+\alpha x\sqrt{2}+\alpha^{2}}+2\arctan\frac{\alpha x\sqrt{2}}{\alpha^{2}-x^{2}}\right) \ \mbox{if} \ AB>0\\ -\frac{1}{4B\alpha'}\left(\ln\frac{x+\alpha'}{x-\alpha'}-2\arctan\frac{x}{\alpha'}\right) \ \mbox{if} \ AB<0 \end{array}\right$$

,where

$$\alpha=\sqrt[4]{\frac{A}{B}}$$

$$\alpha'=\sqrt[4]{\frac{-A}{B}}$$.

Daniel.

-------------------------------------------------
[1]G & R,5-th edition,CD version,Academic Press,1996,formula 2.132,3)

 

[GCT]

alright, I'll try the good o'l fashioned way. Somebody please check my work,

$$I=2 \int \frac{t^{2}~dt}{(1+ \sqrt{2}t+t^{2})(1- \sqrt{2}t +t^{2})}$$

Aside

$$\frac{t^{2}}{(1+ \sqrt{2}t+t^{2})(1- \sqrt{2}t +t^{2})}$$=

$$\frac{Ax+B}{1+ \sqrt{2}t +t^{2}}~+~ \frac{Cx+D}{1- \sqrt{2}t+t^{2}}$$

$$t^{2}=(Ax+B)(1- \sqrt{2}t +t^{2})+(Cx+D)(1+ \sqrt{2}t +t^{2})$$

$$Ax- \sqrt {2} Axt +Axt^{2} +B - \sqrt{2}Bt +Bt^{2} +Cx + \sqrt{2}Cxt +Cxt^{2} +D + \sqrt{2}Dt +Dt^{2}$$

=

$$(Ax+B+Cx+D)t^{2} + (- \sqrt{2} AX - sqrt{2}B + \sqrt{2} Cx + \sqrt{2} D)t +Ax+B+Cx+D$$



1)$$Ax+B+Cx+D=1$$
2)$$- \sqrt{2} Ax - \sqrt{2} B + \sqrt{2} Cx + \sqrt{2}D=0$$
3)$$Ax+B+Cx+D=0$$

It seems that we have a problem here between 1) and 3), it seems to be due to the similarities of the denominator in the original integral...I've never had to deal with a denominator with two different irreducible fractions.

If I did anything wrong, please point it out, or enlighten me toward the appropriate approach.

 

[dextercioby]

Let's stick to the original "t",so the OP could make the connection with the integral which started the thread.

$$\frac{At+B}{1+t\sqrt{2}+t^2}+\frac{Ct+D}{1-t\sqrt{2}+t^2}\equiv \frac{t^{2}}{\mbox{product}}$$

Cross multiplying,one gets the system of equations

$$\left\{\begin{array}{c} A+C=0\\B+D+C\sqrt{2}-A\sqrt{2}=1\\ A+C+D\sqrt{2}-B\sqrt{2}=0\\B+D=0 \end{array}\right$$

which has the unique solution

$$\left\{\begin{array}{cccc} A=-\frac{\sqrt{2}}{4}, & B=0, & C=\frac{\sqrt{2}}{4}, & D=0 \end{array} \right\}$$

Can u carry on from here...?

Daniel.