# Evaluating the indefinite integral $\int \sqrt{\tan x} ~ \mathrm{d}{x}.$

I have been having extreme difficulties with this integral. I would appreciate any and all help.
$$\int \sqrt{\tan x} ~ \mathrm{d}{x}.$$

#### Solutions Collecting From Web of "Evaluating the indefinite integral $\int \sqrt{\tan x} ~ \mathrm{d}{x}.$"

$$y=\int\sqrt{\tan x}\,dx$$
$$g=\int\sqrt{\cot x}\,dx$$
$$y+g=\int\left(\sqrt{\tan x}+\sqrt{\cot x}\right)\,dx$$
$$=\sqrt2\int\frac{\sin x+\cos x}{\sqrt{\sin2x}}\,dx$$
$$=\sqrt2\int\frac{(\sin x-\cos x)’}{\sqrt{1-(\sin x-\cos x)^2}}\,dx$$
$$=\sqrt2\int\frac{u’}{\sqrt{1-u^2}}\,du$$
$$=\sqrt2\sin^{-1}u$$
$$=\sqrt2\sin^{-1}(\sin x-\cos x)$$
$$y-g=\int\left(\sqrt{\tan x}-\sqrt{\cot x}\right)\,dx$$
$$=\sqrt2\int\frac{\sin x-\cos x}{\sqrt{\sin2x}}$$
$$=-\sqrt2\int\frac{(\sin x+\cos x)’}{\sqrt{(\sin x+\cos x)^2-1}}\,dx$$
$$=-\sqrt2\int\frac{s’}{\sqrt{s^2-1}}\,ds$$
$$=-\sqrt2\cosh^{-1}(\sin x+\cos x)$$
$$y=\frac{(y-g)+(y+g)}2$$
$$=\frac{\sqrt2}2(\sin^{-1}(\sin x-\cos x)-\cosh^{-1}(\sin x+\cos x))$$

Let $I = \sqrt{\tan x}\;\mathrm{d}x$ and $J = \sqrt{\cot x}\;\mathrm{d}x$.

Now \begin{align}I + J &= \int\left(\sqrt{\tan x} + \sqrt{\cot x}\right) \;\mathrm{d}x \\ &= \sqrt{2} \int\frac{\sin x + \cos x}{\sqrt{\sin 2x}} \;\mathrm{d}x \\[5pt] &= \sqrt{2} \int\frac{(\sin x – \cos x)’}{\sqrt{1-(\sin x – \cos x)^2}} \;\mathrm{d}x \\[5pt] &= \sqrt{2} \sin^{-1}(\sin x – \cos x) + \mathbb{C_1} \tag{1} \\ \end{align}

and \begin{align}I – J &= \int\left(\sqrt{\tan x} – \sqrt{\cot x}\right) \;\mathrm{d}x \\ &= \sqrt{2} \int\frac{(\sin x – \cos x)}{\sqrt{\sin 2x}} \;\mathrm{d}x \\ &= -\sqrt{2} \int\frac{(\sin x + \cos x)’}{\sqrt{(\sin x + \cos x)^2 – 1}} \;\mathrm{d}x \\ &= -\sqrt{2} \ln\left|(\sin x + \cos x) + \sqrt{(\sin x + \cos x)^2 – 1}\right| + \mathbb{C_2} \tag{2} \\ \end{align}

Now, adding $(1)$ and $(2)$:

$$I = \frac{1}{\sqrt{2}} \sin^{-1}(\sin x – \cos x) – \frac{1}{\sqrt{2}} \ln\left|\sin x + \cos x + \sqrt{\sin 2x} \vphantom{x^{x^x}} \right| + \mathbb{C}$$

Let $u = \sqrt{\tan x}$, then $u^2 = \tan x$. Thus $2u\;\mathrm{d}u = \sec^2 x\;\mathrm{d}x = (u^4 + 1)\mathrm{d}x$. Thus $\mathrm{d}x = \dfrac{2u\;\mathrm{d}u}{u^4 + 1}$. So:

$$\int\sqrt{\tan x}\;\mathrm{d}x = \int\frac{2u^2}{u^4+1}\;\mathrm{d}u$$
You can take it from here.

As already mentioned in some answers, let $t^2=\tan x \implies 2tdt=\sec^2x dx$ or $dx=\frac{2t}{t^4+1}$Now, We can easily reach to the final answer as follows $$I=\int \frac{2t^2 dt}{t^4+1}=\int \frac{2 dt}{t^2+\frac{1}{t^2}}=\int \frac{\left(1+\frac{1}{t^2}\right)+\left(1-\frac{1}{t^2}\right) dt}{t^2+\frac{1}{t^2}}$$ $$=\int \frac{\left(1+\frac{1}{t^2}\right) dt}{t^2+\frac{1}{t^2}}+\int \frac{\left(1-\frac{1}{t^2}\right) dt}{t^2+\frac{1}{t^2}}=\int \frac{\left(1+\frac{1}{t^2}\right) dt}{\left(t-\frac{1}{t}\right)^2+2}+\int \frac{\left(1-\frac{1}{t^2}\right) dt}{\left(t+\frac{1}{t}\right)^2-2}$$ $$=\int \frac{\left(1+\frac{1}{t^2}\right) dt}{\left(t-\frac{1}{t}\right)^2+(\sqrt{2})^2}+\int \frac{\left(1-\frac{1}{t^2}\right) dt}{\left(t+\frac{1}{t}\right)^2-(\sqrt{2})^2}$$ $$=\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{t-\frac{1}{t}}{\sqrt{2}}\right)+\frac{1}{2\sqrt{2}}\ln \left(\frac{\left(t+\frac{1}{t}\right)-\sqrt{2}}{\left(t+\frac{1}{t}\right)+\sqrt{2}}\right)+C$$ Now, substituting the value of $t$, we get $$I=\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{\sqrt{\tan x}-\frac{1}{\sqrt{\tan x}}}{\sqrt{2}}\right)+\frac{1}{2\sqrt{2}}\ln\left(\frac{\sqrt{\tan x}+\frac{1}{\sqrt{\tan x}}-\sqrt{2}}{\sqrt{\tan x}+\frac{1}{\sqrt{\tan x}}+\sqrt{2}}\right)+C$$ $$=\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{\sqrt{\tan x}-\sqrt{\cot x}}{\sqrt{2}}\right)+\frac{1}{2\sqrt{2}}\ln\left(\frac{\sqrt{\tan x}+\sqrt{\cot x}-\sqrt{2}}{\sqrt{\tan x}+\sqrt{\cot x}+\sqrt{2}}\right)+C$$ $$=\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{\sin x-\cos x}{\sqrt{\sin 2x}}\right)+\frac{1}{2\sqrt{2}}\ln\left(\frac{\sin x+\cos x-\sqrt{\sin 2x}}{\sin x+\cos x+\sqrt{\sin 2x}}\right)+C$$

A slight improvement: instead of $u^2=\tan\theta$, let $u^2=2\tan\theta$. This gives
$$I=\frac1{\sqrt2}\int \frac{4u^2}{u^4+4}\,du =\frac1{\sqrt2}\int \frac{u}{u^2-2u+2}-\frac{u}{u^2+2u+2}\,du\ .$$
Observe that except for the constant out the front, no surds are involved. Now substitute $v=u-1$ for the first bit and $v=u+1$ for the second bit. You will need to be careful with the algebra, but it’s not all that bad.

Hint:

Let $\sqrt{\tan x}=u$, $\quad \frac{1}{2\sqrt{\tan x}}\sec^2 x dx=du$, $\frac{1+u^4}{2u}\ dx=du$
$$\int \sqrt{\tan x}\ dx=\int u \frac{2u}{1+u^4}\ du=\int \frac{2u^2}{1+u^4}\ du$$
Now, make partial fractions $$\frac{2u^2}{1+u^4}=\frac{u^2}{(u^2+u\sqrt 2+1)(u^2-u\sqrt 2+1)}$$ Carry on to get the answer