Register to reply 
Integral: square root of tan x 
Share this thread: 
#1
May1306, 05:11 PM

P: 24

My class, teacher included, cannot seem to figure out the integral of the square root of tan x. Maybe someone here can help?
thanks, jeremy 


#2
May1306, 06:03 PM

P: 109

why don't you volunteer to get acreditted ?



#3
May1306, 11:05 PM

P: 989

Examine the given references listed in the archive. Reference: http://www.physicsforums.com/showthread.php?t=91866 http://www.physicsforums.com/showthread.php?t=83012 


#4
May1406, 06:14 AM

Sci Advisor
HW Helper
PF Gold
P: 12,016

Integral: square root of tan x
The simplest way is to set u=sqrt(tan(x)); you'll end up with a rational integrand that you may decompose with partial fractions.
(Remember that sec^2(x)=tan^2(x)+1=u^4+1) 


#5
May1406, 06:32 AM

Emeritus
Sci Advisor
PF Gold
P: 9,772

This may be a bit simplistic but why can't you simply do;
[tex]\int \sqrt{\tan x} \;\; dx = \int \tan^{\frac{1}{2}}x \;\; dx[/tex] [tex] = \frac{3}{2}\tan^{\frac{3}{2}} x = \frac{3}{2}\tan x \sqrt{\tan x} [/tex] ~H 


#6
May1406, 07:15 AM

Sci Advisor
HW Helper
PF Gold
P: 12,016

Hmm..because it is wrong perhaps?
(Differentiate your last expression and see if you get your integrand) 


#7
May1406, 07:30 AM

P: 600

Well Hoot, what you have done is considered tan(x)=u and integrated
u^1/2 du .But you haven't changed dx to du.You can do this as [tex]u=tan(x)^{\frac{1}{2}}[/tex] [tex]\frac{du}{dx}=\frac{sec^2(x)}{2\sqrt{tan(x)}}[/tex] and then find du and so the integrand changes. Just follow Orion's thread to see how it is done. We had the exact same question for our final board exams in India. It took me 10 mins of precious time and two pages of trial to finally get to the answer( a very big one mind you).And to think you did it all for 3 marks in a 100 mark paper.Phew! PS:Something wrong with latex? I just can't seem to edit them. PPS:Hoot, even if you are integrating u^(1/2) it would be 2/3u^3/2 


#8
May1406, 07:35 AM

Emeritus
Sci Advisor
PF Gold
P: 9,772

Ahh, dammit! I knew it was too simple. It's a repeat of my xmas exams when I did a very similar thing with secant! Sorry guys!
~H 


#9
May1606, 05:22 AM

Sci Advisor
HW Helper
P: 11,956

It's much more interesting to consider
[itex] \int \sqrt{\sin x} \ dx [/itex] Daniel. 


#10
May1606, 07:11 AM

P: 989

[itex]\int \sqrt{\tan x} \;\; dx = \frac{1}{2 \sqrt{2}} [ 2 \tan^{1} (1  \sqrt{2} \sqrt{\tan x} ) + 2 \tan^{1} ( \sqrt{2} \sqrt{\tan x} + 1 ) + ...[/itex]
[itex]\ln (  \tan (x) + \sqrt{2} \sqrt{\tan x}  1 )  \ln ( \tan x + \sqrt{2} \sqrt{\tan x} + 1 )][/itex] Reference: http://www.physicsforums.com/showthread.php?t=91866 http://www.physicsforums.com/showthread.php?t=83012 


#11
May1606, 12:37 PM

Sci Advisor
HW Helper
P: 3,025




#12
May1606, 01:13 PM

Sci Advisor
HW Helper
PF Gold
P: 12,016

[tex]\int\frac{2u^{2}du}{\sqrt{1u^{4}}}[/tex] 


#13
May1606, 10:19 PM

P: 989

[tex]\int\frac{2u^{2}du}{\sqrt{1u^{4}}} = \frac{2\,{\sqrt{1  u^2}}\,{\sqrt{1 + u^2}}\,\left( \text{EllipticE}(\sin^{1} u,1) + \text{EllipticF}(\sin^{1} u,1) \right) }{{\sqrt{1  u^4}}}[/tex] Arildno, what are you suggesting for [tex]u[/tex]? 


#14
May1706, 12:36 AM

HW Helper
PF Gold
P: 1,961

while were on the topic of integrating expressions that contain square roots of trigonometric functions, I was having a hard time a while ago evaluating this
[tex]\int{\frac{1}{\sqrt{sin x}}dx[/tex] 


#15
May1706, 01:20 AM

HW Helper
P: 2,954




#16
May1706, 01:26 AM

P: 989

[tex]F(zm) = \text{EllipticF}[z,m] = \int_0^z \frac{1}{\sqrt{1  m \sin^2 t}} dt[/tex]
[tex]\int{\frac{1}{\sqrt{\sin x}}dx = \int_0^{\frac{1}{2} \left( \frac{\pi}{2}  x \right)} \frac{1}{\sqrt{1  2 \sin^2 t}} dt = 2\text{EllipticF} \left[ \frac{1}{2} \left( \frac{\pi}{2}  x \right), 2 \right][/tex] Reference: http://functions.wolfram.com/Ellipti.../EllipticF/02/ 


#17
May1706, 05:13 AM

Sci Advisor
HW Helper
PF Gold
P: 12,016

Given a function f with domain D, the function
[tex]G(x)=\int_{x_{0}}^{x}f(y)dy, x_{0}, y, x\in{D}[/tex] is seen to have no larger domain than f. Since the definite integral can't generate any singularities on its own (integration is a "smoothing" process), it is seen that G doesn't have a less domain than f. Thus, G has the same domain as f. 


#18
May1706, 06:27 AM

P: 989

Given that [itex]\sqrt{\tan x}[/itex] is valid in Quadrants I,III then the specific domains for this function are: [tex]D: \left[ 0, \frac{\pi}{2} \right) \; \; \; I[/tex] [tex]D: \left[ \pi, \frac{3 \pi}{2} \right) \; \; \; III[/tex] The third equation component in post #10 is: [tex]\ln (  \tan x + \sqrt{2} \sqrt{\tan x}  1 )[/tex] Placing the component in a point within its own domain produces: [tex]\ln \left(  \tan \frac{\pi}{4} + \sqrt{2} \sqrt{\tan \frac{\pi}{4}}  1 \right) = \ln (  1 + \sqrt{2} \sqrt{1}  1) = \ln ( \sqrt{2}  2)[/tex] Taking the 'sign' of internal component [itex]\ln [sgn(\sqrt{2}  2)][/itex] yields: [tex]\ln (1)[/tex] Reference: http://mathworld.wolfram.com/Singularity.html http://www.physicsforums.com/showpos...7&postcount=10 


Register to reply 
Related Discussions  
Re: Integral involving square root of e^x  Calculus & Beyond Homework  6  
One integral (square root and gaussian peak)  Calculus  13  
Complex contour integral of squareroot  Calculus & Beyond Homework  1  
Integral with square root  Introductory Physics Homework  13  
Square root of i  General Math  17 