# Find $\int\frac{\tan^2(x)}{\sqrt{x}} \, dx$

I can not find the next antiderivative

$$\displaystyle\int\dfrac{\tan^2(x)}{\sqrt{x}}\,dx$$

## My try

\begin{aligned} \int \dfrac{\sec^2(x)-1}{\sqrt{x}}\,dx =\int\dfrac{\sec^2(x)}{\sqrt{x}}\,dx-\int\dfrac{dx}{\sqrt{x}}=\int\dfrac{\sec^2(x)}{\sqrt{x}}\,dx-2\cdot\sqrt{x}+C\end{aligned} Thank you so much.

• Integration by parts comes to my mind. – A---B Aug 28 '17 at 17:54
• What makes you certain this has a closed form solution at all? – mathematics2x2life Aug 28 '17 at 17:55
• Wolfram alpha can't find a solution in standard time nor can integral calculator. – A---B Aug 28 '17 at 17:56
• Oromion: is this the exact form of the question as is given to you? – Kenny Lau Aug 28 '17 at 18:00
• SymPy,Sage,Maxima,Maple,Mathematica,Rubi - can't find.This is the answer for yours question. – Mariusz Iwaniuk Aug 28 '17 at 19:04

I don't think that there is a closed form solution to your question(as said also in the comments).This is my argument:

Let A=$\int \frac{tan^2(x)}{\sqrt x}dx$.Because I want an integration by parts,I write it as:$\int \frac{tan^2(x)}{\sqrt x} x^\prime dx=\frac{xtan^2(x)}{\sqrt x}-2 \int \frac{xtan(x)}{\sqrt x cos^2(x)}dx+\frac{1}{2}A \Leftrightarrow \frac{1}{2}A=\frac{xtan^2(x)}{\sqrt x}-2\int \frac{xtan(x)}{\sqrt x cos^2(x)}dx$

I want to focus now on $\int \frac{xtan(x)}{\sqrt x cos^2(x)}dx$. With the substitution $tan(x)=t$ we have that $\int \frac{xtan(x)}{\sqrt x cos^2(x)}dx=\int \sqrt {tan^{-1}} \cdot tdt$ or in other words $\int \sqrt{x} \cdot tan(x) dx$.

But when you plug the last integral into Wolfram Alpha it says: " no result found in terms of standard mathematical functions"

I know that maybe the "answer" it's long but I wanted to show you why this has no closed form solution.

• Thaks you for your remark. Maybe you have reason (I hope it). I compute in my HP Prime Graphing Calculator and show me a "monster solution" and don't show step for step. Thanks you really. – Oromion Aug 28 '17 at 20:55
• Glad it helped.Don't worry about it,I'm sure it was some sort of unwanted mistake. Also you can write the Taylor series for the function under the integral just to see how it evolves and then integrate term by term. – Adrian Aug 28 '17 at 21:00
• Guys of M.S.X. My teacher's integral calculus helped me today, the number of exercises is 4b. I cannot imagine the solution, you can see it in (goo.gl/Ccdaqk) – Oromion Aug 31 '17 at 18:15

Only a note.

$\displaystyle \int\frac{(\tan x)^2}{\sqrt{x}}dx = \int\frac{1}{\sqrt{x}}(\frac{d}{dx}\tan x)\,dx - \int\frac{1}{\sqrt{x}} = \frac{\tan x}{\sqrt{x}}-\frac{\sqrt{x}}{2}+ \frac{1}{2}\int\frac{\tan x}{\sqrt{x}^3} dx$

with $\enspace\displaystyle\frac{1}{2}\int\frac{\tan x}{\sqrt{x}^3} dx=\int\frac{\tan (t^2)}{t^2} dt$

The question has changed to find a formula for $\enspace\displaystyle\int\frac{\tan (x^2)}{x^2} dx$

for which the Taylor series around $\,0\,$ is $\enspace\displaystyle \sum\limits_{n=1}^\infty \frac{(-1)^{n-1}2^{2n}(2^{2n}-1)B_{2n}}{(2n)!}\frac{x^{4n-3}}{4n-3}\,$ .

I've never seen a closed form for $\enspace\displaystyle\int\frac{\tan (x^2)}{x^2} dx\,$ , maybe it doesn't exist with conventional functions.

$$\int\dfrac{\tan^2x}{\sqrt x}~dx$$

$$=\int\dfrac{\sec^2x-1}{\sqrt x}~dx$$

$$=\int\dfrac{\sec^2x}{\sqrt x}~dx-\int\dfrac{1}{\sqrt x}~dx$$

$$=\int\dfrac{1}{\sqrt x}~d(\tan x)-2\sqrt x$$

$$=\dfrac{\tan x}{\sqrt x}-\int\tan x~d\left(\dfrac{1}{\sqrt x}\right)-2\sqrt x$$

$$=\dfrac{\tan x}{\sqrt x}-2\sqrt x-\int\tan u^2~d\left(\dfrac{1}{u}\right)$$ (Let $$u=\sqrt x$$)

$$=\dfrac{\tan x}{\sqrt x}-2\sqrt x+\int\dfrac{\tan u^2}{u^2}~du$$

$$=\dfrac{\tan x}{\sqrt x}-2\sqrt x+\int\sum\limits_{n=0}^\infty\dfrac{8}{(2n+1)^2\pi^2-4u^4}~du$$ (use Mittag-Leffler Expansion of tangent)

$$=\dfrac{\tan x}{\sqrt x}-2\sqrt x+\int\sum\limits_{n=0}^\infty\dfrac{8}{((2n+1)\pi+2u^2)((2n+1)\pi-2u^2)}~du$$

$$=\dfrac{\tan x}{\sqrt x}-2\sqrt x+\int\sum\limits_{n=0}^\infty\dfrac{4}{(2n+1)\pi((2n+1)\pi+2u^2)}~du+\int\sum\limits_{n=0}^\infty\dfrac{4}{(2n+1)\pi((2n+1)\pi-2u^2)}~du$$

$$=\dfrac{\tan x}{\sqrt x}-2\sqrt x+\sum\limits_{n=0}^\infty\dfrac{2\sqrt2}{(2n+1)\pi\sqrt{(2n+1)\pi}}\tan^{-1}\dfrac{\sqrt2u}{\sqrt{(2n+1)\pi}}+\sum\limits_{n=0}^\infty\dfrac{2\sqrt2}{(2n+1)\pi\sqrt{(2n+1)\pi}}\tanh^{-1}\dfrac{\sqrt2u}{\sqrt{(2n+1)\pi}}+C$$

$$=\dfrac{\tan x}{\sqrt x}-2\sqrt x+\sum\limits_{n=0}^\infty\dfrac{2\sqrt2}{(2n+1)\pi\sqrt{(2n+1)\pi}}\tan^{-1}\dfrac{\sqrt{2x}}{\sqrt{(2n+1)\pi}}+\sum\limits_{n=0}^\infty\dfrac{2\sqrt2}{(2n+1)\pi\sqrt{(2n+1)\pi}}\tanh^{-1}\dfrac{\sqrt{2x}}{\sqrt{(2n+1)\pi}}+C$$