A Hard Integral

[DubWolf 16,901]

I know the answer is out there somewhere, but try to take this integral:

 

∫√tanx dx

 

Discuss it amongst yourselves.

[SCS 7,470]

My initial thought is to use integration by parts.
 
I would set the following:
 
u = √tanx
du = (1/2) * (tan x)^(-1/2) * [(sec x)^2] dx
 
dv = dx
v = x
 
Via the integration of parts formula:

∫√tanx dx = (√tanx) * x - [(1/2) * ∫ x * (tan x)^(-1/2) * [(sec x) ^ 2] ]
 
Unfortunately, my new integrand is more complicated than the original, so initially using integration by parts doesn't work here. A u-substitution would work at this step if there wasn't a factor of x in this integrand. Further iterations of integration by parts would be a fruitless endeavor as that will only continue to increase the complexity of the integrand due to the nature of what this step ascertained.
 
After exhausting that method, I'm pretty much at a loss as to how to solve this problem.
 
This is one of the many things I love about integration. A function such as the square root of tan x might look deceptively simple at first, but integrating it is astoundingly difficult. Yet, a function such as [2*(ln x)^3] / x, despite the fact that at an initial glance it looks more complicated, would be very simple to integrate. First, recognize that the function can be rewritten in an equivalent form as: [2*(ln x)*[(ln x) ^ 2]]/x. Then, you would just set u = (ln x) ^ 2 and du = 2*(ln x) / x dx. The integrand would then be u du, with a simple antiderivative of (u^2)/2 + C. Substituting back in for u, the antiderivative is [(ln x) ^ 4] / 2 + C.

Going back to the original problem of integrating the square root of tan x: I looked up a step-by-step solution online and I understand how it works, but it's not something I would have thought of on my own.
 
PM me if you want the link to it, but I'll hold off on posting it here for awhile to give other people more time to try to figure it out.

[Twilight Sparkle ✨ 8,522]

Integrals involving trig functions sometimes "loop around" when you take the integral by parts a few times and pull the original integral out again, at which point you can usually solve it algebraically. Is this one of those integrals?

[SCS 7,470]

Integrals involving trig functions sometimes "loop around" when you take the integral by parts a few times and pull the original integral out again, at which point you can usually solve it algebraically. Is this one of those integrals?

 

I think that's most likely to happen with integrands involving sine and cosine due to the nature of their derivatives and subsequently their antiderivatives. Other trigonometric functions generally have more complicated derivatives and antiderivatives so you would be buried in layers of algebra making the problem virtually impossible before anything would loop around much if at all. I'm not saying it's necessarily impossible: it would likely depend on the choices you made in regard to parts/substitutions.

 

As this problem involves the tangent function, I would be inclined to say that this is not an integral that could loop easily if at all.

[Guest90210 3,567]

Uh...I can do addition, and subtraction...Sometimes I can divide or multiply if the numbers are small...So I'll give it a shot.

 

∫√tanx dx

 

Well, tan has 3 letters. DX was a wrestling group comprised of Triple H and Shawn Michaels. Add that extra X to the end of DX, and you have DXX. X is the 24th letter of the alphabet. Multiply that by 2, and you get 48. D = 4. 48 divided by 4 is 12. 12 - 2 (the amount of members in the last surviving version of DX) and you get 10.

 

The answer is 10.   

Edited October 2, 2014 by Rivendare

[Jeric 46,838]

Yeah hold off SCS, I'll tackle this when I get home. Been a while so this should be fun (or a train wreck)

[Fhaolan 4,471]

Yeah, that's a bloody mess, that is.

 

Let's see.. u substitution on √tan(x) means du = 1 / (2cos²(x)√tan(x)

 

But how do you get factor √tan(x) to have 1 / (2cos²(x)√tan(x) in it somewhere, while leaving all the other factors in terms that resemble √tan(x)?

 

....

 

 

It's going to have something to do with one of the trig identities, I'm sure of it. But it will be some twisted version that you'll only recognize if you've been doing trig integration for awhile.