Integral of $\sqrt{\tan x}$

Question

Evaluate this integral: $\int\sqrt{\tan x } \ dx$

So I had an idea, which just looks too simple to be true, but correct me please, I believe it has to be wrong...

1. Rewrite as: $$\int\sqrt\frac{\sin x}{\cos x} dx$$

2. Make substitutions, and work around the terms: $$\sqrt{\cos x}=t$$ $$dt = -\frac{\sin x}{2\sqrt{\cos x}}dx$$ Now you have to look at the original integral and notice $\frac{dx}{\sqrt{\cos x}}$ is there, so work it around and get the sub.

3. Rewrite the original term in terms of $dt$.$$-\int\frac{\sqrt{\sin x}}{\sin x}2dt$$

4. See? Now intuitively, just do the integral which is very easy, and return the substituted $t$. The result is: $$-2\frac{\sqrt{\cos x}}{\sqrt{\sin x}}+C$$

To me, it seems wrong when I compare it to the results around the internet, but none of these operations seems illegal to me :)).

So what do you say?

Answer 1

Try: $$\int\sqrt{\tan x}dx=\int\frac{2u^2}{u^4+1}du=2\int{\frac{{u}^{2}}{\left({u}^{2}-\sqrt{2}u+1\right)\left({u}^{2}+\sqrt{2}u+1\right)}}du=$$ $$2 \int{\frac{u}{2\sqrt{2}\left({u}^{2}-\sqrt{2}u+1\right)}-\frac{u}{2\sqrt{2}\left({u}^{2}+\sqrt{2}u+1\right)}}du=$$ $$=-\frac{\ln\left({u}^{2}+\sqrt{2}u+1\right)}{2\sqrt{2}}+\frac{\ln\left({u}^{2}-\sqrt{2}\,u+1\right)}{2\sqrt{2}}+\frac{\arctan\left(\frac{2\,u+\sqrt{2}}{\sqrt{2}}\right)}{\sqrt{2}}+\frac{\arctan\left(\frac{2u-\sqrt{2}}{\sqrt{2}}\right)}{\sqrt{2}}+C$$

Answer 2

Let $$I=\int\sqrt{\tan(x)}dx$$

Let $\tan(x)=u^{2}$

$$\implies \sec^{2}(x)dx=2udu$$

$$\implies dx=\frac{2u}{1+u^{4}}du$$

$$\implies I=\int \frac{u(2u)}{1+u^{4}}du$$

$$\implies I=\int \frac{2u^{2}}{1+u^{4}}du$$

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

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

Now dividing the numerator and denominator by $u^{2}$ we get:

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

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

Now for the $1^{st}$ part substitute $(u-\frac{1}{u})=t$ and then $(1+\frac{1}{u^{2}})du=dt$. Again for the $2^{nd}$ part substitute $(u+\frac{1}{u})=z$ and then $(1-\frac{1}{u^{2}})du=dz$.

$$\implies I=\int\frac{1}{t^{2}+2}dt +\int\frac{1}{z^{2}-2}dz$$

Now try to solve. Enough hints are given.