Is there a simpler method for finding the integral of sqrt(tan x)?

[murshid_islam]

My question is about this integral:

\int\sqrt{\tan (x)}dx

After using the substitution, u² = tan(x), I got,

2\int\frac{u^2}{u^4 + 1}du = 2\int\frac{u^2}{\left(u^2 + \sqrt{2}u + 1\right)\left(u^2 - \sqrt{2}u + 1\right)}du

Next, I tried the partial fraction expansion. But it turned pretty ugly. Is there any easier way of doing it?

 

[nicksauce]

If it is a definite integral, I think contour integration would be the way to go... otherwise I can't think of a way to make it nice.

 

[dextercioby]

I don't think there's a neater way to compute it other than the method you already have.

 

[Gib Z]

There is;

\int \frac{2u^2}{u^4+1} du = \int \frac{u^2+1}{u^4+1} du + \int \frac{u^2-1}{u^4+1} du

= \int \frac{1+\frac{1}{u^2}}{u^2+\frac{1}{u^2}} du + \int \frac{1-\frac{1}{u^2}}{u^2+\frac{1}{u^2}} du

= \int \frac{ d(u-\frac{1}{u})}{ (u-\frac{1}{u})^2 +2} du + \int \frac{ d(u+\frac{1}{u})}{ (u+\frac{1}{u})^2 -2} du

 

[murshid_islam]

There is;

\int \frac{2u^2}{u^4+1} du = \int \frac{u^2+1}{u^4+1} du + \int \frac{u^2-1}{u^4+1} du

= \int \frac{1+\frac{1}{u^2}}{u^2+\frac{1}{u^2}} du + \int \frac{1-\frac{1}{u^2}}{u^2+\frac{1}{u^2}} du

= \int \frac{ d(u-\frac{1}{u})}{ (u-\frac{1}{u})^2 +2} du + \int \frac{ d(u+\frac{1}{u})}{ (u+\frac{1}{u})^2 -2} du

I'm sorry, but I don't see exactly how that helps.

 

[Gib Z]

The remaining integrals are of forms which are generally considered standard integrals;

\int \frac{dx}{x^2+a^2} = \frac{1}{a} \tan^{-1} \frac{x}{a} + C

\int \frac{dx}{x^2-a^2} = \frac{1}{2a} \log_e \left( \frac{x-a}{x+a}\right) + C

 

[murshid_islam]

But how do I deal with the d\left(u-\frac{1}{u}\right) and d\left(u+\frac{1}{u}\right) in the numerators of the integrals?

 

[Count Iblis]

Substitution: put z = u-1/u in the first integral and z = u+1/u in the second integral

 

[TheoMcCloskey]

murshid_islam - Just to be clear, the trailing du in Gib Z's brillant integrals should be deleted. The form of the integrals are really

\int \frac{2u^2}{u^4+1} du = \int \frac{ d(u-\frac{1}{u})}{ (u-\frac{1}{u})^2 +2} + \int \frac{ d(u+\frac{1}{u})}{ (u+\frac{1}{u})^2 -2}

Do you now understand how the substitution gets to the integral forms in his second post?

 

[murshid_islam]

murshid_islam - Just to be clear, the trailing du in Gib Z's brillant integrals should be deleted.

Thanks a lot. Now everything is clear to me.

Do you now understand how the substitution gets to the integral forms in his second post?

Oh yes, now I understand.

 

[Gib Z]

Argh! All that latex code makes me forget what I'm actually typing and it's just instict to jab a du in there after an integral! Sorry murshid_islam!