1. May 31, 2005

$$\int{\sqrt{\tanx}dx}[\tex] please help me as soon as possible because i want to submit my assignment tomorrow 2. May 31, 2005 ### whozum [tex]\int{\sqrt{\tan x}dx}$$

I would imagine if you played around with it enough you could get it by parts. I cant see anything obvious though.

3. May 31, 2005

### Oggy

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

4. May 31, 2005

### 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 [Broken]

Last edited by a moderator: May 2, 2017
5. May 31, 2005

### 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.

6. May 31, 2005

### whozum

Oh, ofcourse. How could I forget that.

7. May 31, 2005

### PhysicsinCalifornia

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 ?

Last edited: May 31, 2005
8. May 31, 2005

### whozum

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

9. May 31, 2005

### shaiqbashir

Thanks a lot for ur precious help!

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

10. May 31, 2005

### 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

#### Attached Files:

• ###### Integrate.gif
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11. May 31, 2005

### Oggy

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

12. May 31, 2005

### whozum

Oggy the final integral is

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

13. May 31, 2005

### whozum

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

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

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

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

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

Last edited: May 31, 2005
14. Jun 1, 2005

### Oggy

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

15. Jun 1, 2005

### 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 im 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.

16. Jun 1, 2005

### 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.

17. Jun 1, 2005

### 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?

18. Jun 1, 2005

### 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.:tongue2:

Daniel.

19. Jun 1, 2005

### calculus1967

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

20. Jun 1, 2005

### 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 $-\cos^2 x$