# Integral: square root of tan x

by Jeremy
Tags: integral, root, square
 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
 P: 109 why don't you volunteer to get acreditted ?
 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
 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)
 Emeritus Sci Advisor PF Gold P: 9,772 This may be a bit simplistic but why can't you simply do; $$\int \sqrt{\tan x} \;\; dx = \int \tan^{\frac{1}{2}}x \;\; dx$$ $$= \frac{3}{2}\tan^{\frac{3}{2}} x = \frac{3}{2}\tan x \sqrt{\tan x}$$ ~H
 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)
 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 $$u=tan(x)^{\frac{1}{2}}$$ $$\frac{du}{dx}=\frac{sec^2(x)}{2\sqrt{tan(x)}}$$ 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
 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
 Sci Advisor HW Helper P: 11,956 It's much more interesting to consider $\int \sqrt{\sin x} \ dx$ Daniel.
 P: 989 $\int \sqrt{\tan x} \;\; dx = \frac{1}{2 \sqrt{2}} [ 2 \tan^{-1} (1 - \sqrt{2} \sqrt{\tan x} ) + 2 \tan^{-1} ( \sqrt{2} \sqrt{\tan x} + 1 ) + ...$ $\ln (| - \tan (x) + \sqrt{2} \sqrt{\tan x} - 1 |) - \ln (| \tan x + \sqrt{2} \sqrt{\tan x} + 1 |)]$ Reference: http://www.physicsforums.com/showthread.php?t=91866 http://www.physicsforums.com/showthread.php?t=83012
HW Helper
P: 3,025
 Quote by dextercioby It's much more interesting to consider $\int \sqrt{\sin x} \ dx$ Daniel.
HW Helper
PF Gold
P: 12,016
 Quote by dextercioby It's much more interesting to consider $\int \sqrt{\sin x} \ dx$ Daniel.
$$\int\frac{2u^{2}du}{\sqrt{1-u^{4}}}$$

P: 989
 Quote by arildo $$\int\frac{2u^{2}du}{\sqrt{1-u^{4}}}$$

$$\int\frac{2u^{2}du}{\sqrt{1-u^{4}}} = \frac{-2\,{\sqrt{1 - u^2}}\,{\sqrt{1 + u^2}}\,\left( -\text{EllipticE}(\sin^{-1} u,-1) + \text{EllipticF}(\sin^{-1} u,-1) \right) }{{\sqrt{1 - u^4}}}$$

Arildno, what are you suggesting for $$u$$?
 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 $$\int{\frac{1}{\sqrt{sin x}}dx$$
HW Helper
P: 2,954
 Quote by dx 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 $$\int{\frac{1}{\sqrt{sin x}}dx$$
It is always a good idea when confronted with an unfamiliar integral, to verify that it can be done before expending effort to figure out how. Mathematica is a good tool, or you use the free WebMathematica equivalent at http://integrals.wolfram.com/index.jsp
 P: 989 $$F(z|m) = \text{EllipticF}[z,m] = \int_0^z \frac{1}{\sqrt{1 - m \sin^2 t}} dt$$ $$\int{\frac{1}{\sqrt{\sin x}}dx = \int_0^{\frac{1}{2} \left( \frac{\pi}{2} - x \right)} \frac{1}{\sqrt{1 - 2 \sin^2 t}} dt = -2\text{EllipticF} \left[ \frac{1}{2} \left( \frac{\pi}{2} - x \right), 2 \right]$$ Reference: http://functions.wolfram.com/Ellipti.../EllipticF/02/
 Sci Advisor HW Helper PF Gold P: 12,016 Given a function f with domain D, the function $$G(x)=\int_{x_{0}}^{x}f(y)dy, x_{0}, y, x\in{D}$$ 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.
 P: 989 Given that $\sqrt{\tan x}$ is valid in Quadrants I,III then the specific domains for this function are: $$D: \left[ 0, \frac{\pi}{2} \right) \; \; \; I$$ $$D: \left[ \pi, \frac{3 \pi}{2} \right) \; \; \; III$$ The third equation component in post #10 is: $$\ln ( - \tan x + \sqrt{2} \sqrt{\tan x} - 1 )$$ Placing the component in a point within its own domain produces: $$\ln \left( - \tan \frac{\pi}{4} + \sqrt{2} \sqrt{\tan \frac{\pi}{4}} - 1 \right) = \ln ( - 1 + \sqrt{2} \sqrt{1} - 1) = \ln ( \sqrt{2} - 2)$$ Taking the 'sign' of internal component $\ln [sgn(\sqrt{2} - 2)]$ yields: $$\ln (-1)$$ Reference: http://mathworld.wolfram.com/Singularity.html http://www.physicsforums.com/showpos...7&postcount=10

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