# Integrating the following $\int \sqrt{\tan x+1}\,dx$

Question: Integrate the following, $$\int\sqrt{\tan x+1}\;dx.$$

First consider this integral:

$$\int \frac{1}{(x+1)\sqrt{x+3}} \, dx = -\sqrt{2}\tanh^{-1}\frac{\sqrt{x+3}}{\sqrt{2}} + c$$

Wolfram Alpha confirms that result.

Then, we have

$$I=\int \sqrt{\tan x+1} \, dx, \quad \tan x=u+2, \quad dx=\frac{du}{\sec^{2}x}=\frac{du}{(u+2)^{2}-1}=\frac{dx}{(u+3)(u+1)}$$

So this transforms the integral to the first integral on this post, which we can evaluate. Then after evaluation and resubstitution I get:

$$I=-\sqrt{2}\tanh^{-1}\frac{\sqrt{\tan x+1}}{\sqrt{2}}+c$$

However differentiating this with Wolfram Alpha gives me a messy trigonometric expression which doesn't seem to be equal (I tested some values in both expressions and get different answers). I also estimated the area under the integral between some values and also obtained different answers using the closed form. Any ideas why?

EDIT: I used the wrong identity. Nevertheless, we can still use this method to integrate sqrt(tanhx integrals). E.g: $$I=\int \sqrt{\tanh x+1} \, dx, \tanh x=u+2,\quad -dx = \frac{du}{\operatorname{sech}^2 x} = \frac{du}{(u+2)^2-1} = \frac{dx}{(u+3)(u+1)}$$

To obtain: $\int \sqrt{\tanh x+1} \, dx = I=\sqrt{2}\tanh^{-1} \dfrac{\sqrt{\tanh x+1}}{\sqrt{2}}+c$

• $\sec^2x=1+\tan^2x=1+(u+2)^2$. – egreg Jun 4 '16 at 13:12
• $\sec^2(x)=\tan^2(x)\color{red}{+}1$ and not $\color{green}{-}$ as you have. Exactly as @egreg said – b00n heT Jun 4 '16 at 13:12
• Oh yes, my bad. Nevertheless this was not in vain as it can be used to integrate sqrt(tanhx+1) – Laksh Jun 4 '16 at 13:14
• Similar to math.stackexchange.com/questions/626942 – Harry Peter Jun 10 '16 at 15:32

Hint:

Let $u=\sqrt{\tan x+1}$ ,

Then $x=\tan^{-1}(u^2-1)$

$dx=\dfrac{2u}{(u^2-1)^2+1}du$

$\therefore\int\sqrt{\tan x+1}~dx=\int\dfrac{2u^2}{(u^2-1)^2+1}du$

• Good manipulation, however the following integral still cannot be expressed in terms of elementary functions (you could factorise the quartic denominator into complex factors and thus have a closed form of the integral with complex numbers). – Laksh Jun 15 '16 at 10:16
• That's not true. Wolfram Alpha gives the result in terms of the inverse hyperbolic tangent, which it not non-elementary. – Dylan Dec 3 '17 at 5:38

directly use:$$u=\sqrt{\tan x+1}$$ it's possible to solve with that

This integral is not non-elementary. Following from @Harry Peter's hint, we have

\begin{align} \frac{2u^2}{u^4 - 2u^2 + 2} &= \frac{2}{u^2 + \dfrac{2}{u^2} - 2} \\ &= \frac{1 - \dfrac{\sqrt{2}}{u^2}}{\left(u+\dfrac{\sqrt{2}}{u}\right)^2-2\sqrt{2}-2} + \frac{1 + \dfrac{\sqrt{2}}{u^2}}{\left(u - \dfrac{\sqrt{2}}{u}\right)^2 + 2\sqrt{2} - 2 } \end{align}

Then you can substitute $s = u + \dfrac{\sqrt{2}}{u}$ and $t = u - \dfrac{\sqrt{2}}{u}$, respectively. The first integral is in terms of logarithms and the second is in terms of arctangent. As you can see, this can be expressed in terms of elementary functions.

$$I=\int\sqrt{\tan x+1}\ dx$$ The substitution $$u=\sqrt{\tan x+1}$$ gives $$I=2\int\frac{u^2du}{(u^2-1)^2+1}$$

A (not) Quick Detour:

Now we're gonna do a sneaky fraction decomposition. If $$H(x)$$ is a polynomial of degree $$n$$, then it has $$n$$ roots. If the roots of $$H(x)$$ are non-repeating, then $$H(x)$$ can be factored as follows: $$H(x)=F\prod_{r\in Q}(x-r)$$ Where $$Q=\{x:H(x)=0\}$$ is the set of roots of $$H(x)$$, and $$F$$ is some constant. And $$\prod_{i=1}^ka_i:=a_1\cdot a_2\cdots a_k=\prod_{i\in\{1,2,...,k\}}a_i$$ Because of this we know that $$\frac1{H(x)}$$ can be represented like so, as a decomposed fraction: $$\frac1{H(x)}=\frac1F\sum_{r\in Q}\frac{b(r)}{x-r}$$ Where $$b(r)$$ represents the constants that come as a result of the fraction decomposition. These constants are independent of $$x$$. Here's how we do the thing: $$\frac1{H(x)}=\frac1F\sum_{r\in Q}\frac{b(r)}{x-r}$$ $$\frac1{F\prod_{r\in Q}(x-r)}=\frac1F\sum_{r\in Q}\frac{b(r)}{x-r}$$ $$\prod_{r\in Q}\frac1{x-r}=\sum_{r\in Q}\frac{b(r)}{x-r}$$ $$\bigg(\prod_{a\in Q}(x-a)\bigg)\bigg(\prod_{r\in Q}\frac1{x-r}\bigg)=\bigg(\prod_{a\in Q}(x-a)\bigg)\sum_{r\in Q}\frac{b(r)}{x-r}$$ $$1=\sum_{r\in Q}\frac{b(r)}{x-r}\prod_{a\in Q}(x-a)$$ $$1=\sum_{r\in Q}b(r)\prod_{r\neq a\in Q}(x-a)$$ Thus, for any $$k\in Q$$, $$1=\sum_{r\in Q}b(r)\prod_{r\neq a\in Q}(k-a)$$ $$1=b(k)\prod_{k\neq a\in Q}(k-a)$$ $$b(k)=\prod_{k\neq a\in Q}\frac1{k-a}$$ Which of course gives: $$\frac1{H(x)}=\frac1F\sum_{r\in Q}\frac1{x-r}\prod_{r\neq a\in Q}\frac1{r-a}$$ Which is integrated easily.

Again this can only be done with $$H(x)$$ if it's roots are non-repeating, because they weren't non-repeating, there would exist a pair of distinct $$r_1,r_2\in Q$$, were $$r_1=r_2$$, which would make the $$\prod_{r\neq a\in Q}\frac1{r-a}$$ bit become $$\frac10$$ in the case that $$r=r_1$$, and $$a=r_2$$. That was probably worded horribly, but just take my word for it that the roots of $$H(x)$$ must be non-repeating.

Back to the Integral:

We may choose $$H(u)=(u^2-1)^2+1$$, which has $$4$$ (non-repeating: yay!) roots all in the form $$u=\pm\sqrt{1\pm i}$$ Thus we have, when $$R=\{u:(u^2-1)^2+1=0\}$$, $$I=2\int u^2\sum_{r\in R}\frac1{u-r}\prod_{r\neq a\in R}\frac1{r-a}\ du$$ Which simplifies: $$I=2\sum_{r\in R}\prod_{r\neq a\in R}\frac1{r-a}\int\frac{u^2du}{u-r}$$

Another Detour:

Consider now $$K=\int\frac{x^2dx}{x-r}$$ The substitution $$w=x-r$$ gives $$K=\int\frac{(w+r)^2 dw}w$$ $$K=\int\frac{w^2+2rw+r^2}w dw$$ $$K=\int\frac{w^2}w dw+2r\int\frac{w}wdw+r^2\int\frac1wdw$$ $$K=\frac{w^2}2+2rw+r^2\log|w|$$ $$K=\frac{(x-r)^2}2+2r(x-r)+r^2\log|x-r|$$ $$K=\frac{(x-r)(x-3r)}2+r^2\log|x-r|$$

The final bit:

Plugging in $$K$$ gives $$I=2\sum_{r\in R}\bigg(r^2\log|u-r|+\frac{(u-r)(u-3r)}2\bigg)\prod_{r\neq a\in R}\frac1{r-a}$$ $$I=\sum_{r\in R}\big(2r^2\log|u-r|+(u-r)(u-3r)\big)\prod_{r\neq a\in R}\frac1{r-a}$$ $$I=\sum_{r\in R}\frac{2r^2\log|u-r|+(u-r)(u-3r)}{\prod_{r\neq a\in R}(r-a)}$$

Don't forget to plug in $$u=\sqrt{\tan x+1}$$. :)