Question

What's the anti-derivative of the square root of tangent x? Show all steps

Answer 1

\[\int\limits \sqrt{\tan x}dx\]?

Answer 2

Yes Alex

Answer 3

that is:\[(\tan(x))^{1/2}\]

Answer 4

now you can set u=tan(x):
so:
\[\int\limits_{}^{}u^{1/2}\]
equals:
\[\frac{ 2 }{ 3 }u^{3/2}\]

Answer 5

now sub back in tan(x) for u

Answer 6

so you get:\[\frac{ 2 }{ 3 }(\tan(x))^{3/2}\]

Answer 7

tanx^3/2 equals:\[\sqrt{(\tan(x))^3}\]

Answer 8

so the whole thing is;
\[\frac{ 2 }{ 3 }\sqrt{(\tan(x))^3}\]

Answer 9

or you can write it as:\[\frac{ 2\sqrt{(\tan(x))^3} }{ 3 }\]

Answer 10

I'm confused. Can you please show me the steps? The last thing you wrote isn't the answer in my book.

Answer 11

I'm not sure about that integral?

Answer 12

This is going to be fun. XD

Let \(\large \color{green}{t^2}=\color{green}{\tan x}\implies \color{red}{\sec^2x}=\tan^2x+1 = \color{red}{t^4+1}\). Also note that
\[\large \begin{aligned} t^2=\tan x &\implies 2t\,dt = \color{red}{\sec^2x}\,dx\\ & \implies 2t\,dt = (\color{red}{t^4+1})\,dx \\ &\implies \color{blue}{\frac{2t}{t^4+1}\,dt} = \color{blue}{\,dx}.\end{aligned}\]Therefore
\[\large \int\sqrt{\color{green}{\tan x}}\color{blue}{\,dx} \xrightarrow{t^2=\tan x}{} \int\frac{\sqrt{\color{green}{t^2}}\cdot \color{blue}{2t}}{\color{blue}{t^4+1}}\color{blue}{\,dt} = 2\int \frac{t^2}{t^4+1}\,dt\]Now, this is where things start to get fun. We first rewrite \(\large t^4+1\) by noting that
\[\large \begin{aligned}t^4+1 &= t^4+2t^2+1 - 2t^2\\ &= (t^2+1)^2-(\sqrt{2}t)^2\\&= (t^2+1+\sqrt{2}t)(t^2+1-\sqrt{2}t) \\ &= (t^2+\sqrt{2}t+1)(t^2-\sqrt{2}t+1)\end{aligned}\]We now use partial fraction to find \(A,B,C,D\) so that
\[\large\frac{t^2}{\underbrace{(t^2+\sqrt{2}t+1)(t^2-\sqrt{2}t+1)}_{t^4+1}} = \frac{At+B}{t^2+\sqrt{2}t+1}+\frac{Ct+D}{t^2-\sqrt{2}t+1}\]Multiplying both sides by the common denominator gives us
\[\large \begin{aligned} t^2&= (At+B)(t^2-\sqrt{2}t+1)+(Ct+D)(t^2+\sqrt{2}t+1)\\ &= (A+C)t^3+(B-\sqrt{2}A+D+\sqrt{2}C)t^2 \\ & \phantom{= (A+C)t^3+(B-\sqrt{2}A}+ (A-\sqrt{2}B+C+\sqrt{2}D)t + B+D \end{aligned}\]Comparing coefficients of LHS and RHS yields the system of equations
\[\large \left\{\begin{aligned}A+C &= 0\\ B-\sqrt{2}A+D+\sqrt{2}C &= 1\\ A-\sqrt{2}B+C+\sqrt{2}D &= 0\\ B+D &= 0\end{aligned}\right.\]So far, we see that \(\large A=-C\) and \(\large B=-D\). Substituting these two into the second and third equations leaves us with
\[\large\left\{\begin{aligned}-D+\sqrt{2}C+D+\sqrt{2}C &= 1 \\ -C+\sqrt{2}D+C+\sqrt{2}D &= 0 \end{aligned}\right. \implies \left\{\begin{aligned}2\sqrt{2}C &= 1 \\ 2\sqrt{2}D &= 0 \end{aligned}\right.\]Hence, \(\large C=\dfrac{1}{2\sqrt{2}}\implies A=-\dfrac{1}{2\sqrt{2}}\) and \(\large D=0\implies B=0\). Thus,
\[ \frac{t^2}{\underbrace{(t^2+\sqrt{2}t+1)(t^2-\sqrt{2}t+1)}_{t^4+1}} = \frac{1}{2\sqrt{2}}\left[\frac{-t}{t^2+\sqrt{2}t+1} + \frac{t}{t^2-\sqrt{2}t+1}\right]\]Therefore, we substitute the partial fraction into the integral (and complete the square) to get
\[\large \begin{aligned}&\phantom{=}2\int\frac{t^2}{t^4+1}\,dt\\ &= \frac{2}{2\sqrt{2}}\int \left[\frac{-t}{t^2+\sqrt{2}t+1} + \frac{t}{t^2-\sqrt{2}t+1}\right]\,dt\\ &=\frac{\sqrt{2}}{2} \left[-\int\frac{t}{\left(t+\frac{\sqrt{2}}{2} \right)^2+\frac{1}{2}}\,dt + \int\frac{t}{\left(t-\frac{\sqrt{2}}{2}\right)^2+\frac{1}{2}}\,dt\right]\\ &= \frac{\sqrt{2}}{2}\left[-\int\frac{\left(t+\frac{\sqrt{2}}{2}\right) - \frac{\sqrt{2}}{2}}{\left(t+\frac{\sqrt{2}}{2} \right)^2+\frac{1}{2}}\,dt + \int\frac{\left(t-\frac{\sqrt{2}}{2}\right)+\frac{\sqrt{2}}{2}}{\left(t-\frac{\sqrt{2}}{2}\right)^2+\frac{1}{2}}\,dt\right]\\ &= \frac{\sqrt{2}}{2}\left[-\int\frac{t+\frac{\sqrt{2}}{2}}{\left(t+\frac{\sqrt{2}}{2} \right)^2 +\frac{1}{2}}\,dt +\frac{\sqrt{2}}{2} \int\frac{1}{\left(t+\frac{\sqrt{2}}{2}\right)^2+\frac{1}{2}}\,dt\right.\\ &\phantom{=\frac{\sqrt{2}}{2}x.} \left.+\int\frac{t-\frac{\sqrt{2}}{2}}{\left(t-\frac{\sqrt{2}}{2}\right)^2+\frac{1}{2}}\,dt +\frac{\sqrt{2}}{2} \int\frac{1}{\left(t-\frac{\sqrt{2}}{2}\right)^2+\frac{1}{2}}\,dt\right]\\ &= \frac{\sqrt{2}}{2}\left[\underbrace{-\int\frac{t+\frac{\sqrt{2}}{2}}{\left(t+\frac{\sqrt{2}}{2} \right)^2 +\frac{1}{2}}\,dt}_{I_1} +\underbrace{\int\frac{\sqrt{2}}{\left(\sqrt{2}t+1\right)^2+1}\,dt}_{I_2} \right.\\ &\phantom{=\frac{\sqrt{2}}{2}x.} \left.+\underbrace{\int\frac{t-\frac{\sqrt{2}}{2}}{\left(t-\frac{\sqrt{2}}{2}\right)^2+\frac{1}{2}}\,dt}_{I_3} + \underbrace{\int\frac{\sqrt{2}}{\left(\sqrt{2}t-1\right)^2+1 }\,dt}_{I_4}\right] \end{aligned}\]Now, consider each of these integrals separately.

----------

\[\large I_1 = -\int \frac{t+\frac{\sqrt{2}}{2}}{\left(t+\frac{\sqrt{2}}{2}\right)^2+\frac{1}{2}}\,dt\]Let \(\large s=\left(t+\frac{\sqrt{2}}{2}\right)^2+\frac{1}{2} \implies \,ds = 2\left(t+\frac{\sqrt{2}}{2}\right)\,dt\). Thus,
\[\large \begin{aligned}I_1 = -\int \frac{t+\frac{\sqrt{2}}{2}}{\left(t+\frac{\sqrt{2}}{2}\right)^2+\frac{1}{2}}\,dt \xrightarrow{s}{} &\phantom{=} -\frac{1}{2}\int\frac{1}{s}\,ds \\ &= -\frac{1}{2}\ln|s| + C\\ &= -\frac{1}{2}\ln\left(\left(t+\tfrac{\sqrt{2}}{2}\right)^2+\tfrac{1}{2}\right)+C \\ &= -\frac{1}{2}\ln(t^2+\sqrt{2}t+1)+C\end{aligned}\]
----------

\[\large I_2 = \int \frac{\sqrt{2}}{\left(\sqrt{2}t+1\right)^2+1}\,dt\]let \(\large s=\sqrt{2}t+1 \implies \,ds = \sqrt{2} \,dt\). Thus,
\[\large \begin{aligned}I_2 = \int \frac{\sqrt{2}}{\left(\sqrt{2}t+1\right)^2+1}\,dt \xrightarrow{s=\sqrt{2}t+1}{} &\phantom{=} \int\frac{1}{s^2+1}\,ds \\ &= \arctan s + C\\ &= \arctan(\sqrt{2}t+1)+C\end{aligned}\]
----------

\[\large I_3 = \int \frac{t-\frac{\sqrt{2}}{2}}{\left(t-\frac{\sqrt{2}}{2}\right)^2+\frac{1}{2}}\,dt\]Let \(\large s=\left(t-\frac{\sqrt{2}}{2}\right)^2+\frac{1}{2} \implies \,ds = 2\left(t-\frac{\sqrt{2}}{2}\right)\,dt\). Thus,
\[\large \begin{aligned}I_3 = \int \frac{t-\frac{\sqrt{2}}{2}}{\left(t-\frac{\sqrt{2}}{2}\right)^2+\frac{1}{2}}\,dt \xrightarrow{s}{} &\phantom{=} \frac{1}{2}\int\frac{1}{s}\,ds \\ &= \frac{1}{2}\ln|s| + C\\ &= \frac{1}{2}\ln\left(\left(t-\tfrac{\sqrt{2}}{2}\right)^2+\tfrac{1}{2}\right)+C \\ &= \frac{1}{2}\ln(t^2-\sqrt{2}t+1)+C\end{aligned}\]
----------

\[\large I_4 = \int \frac{\sqrt{2}}{\left(\sqrt{2}t-1\right)^2+1}\,dt\]let \(\large s=\sqrt{2}t-1 \implies \,ds = \sqrt{2} \,dt\). Thus,
\[\large \begin{aligned}I_4 = \int \frac{\sqrt{2}}{\left(\sqrt{2}t-1\right)^2+1}\,dt \xrightarrow{s=\sqrt{2}t-1}{} &\phantom{=} \int\frac{1}{s^2+1}\,ds \\ &= \arctan s + C\\ &= \arctan(\sqrt{2}t-1)+C\end{aligned}\]Thus, we see that
\[\begin{aligned}2\int\frac{t^2}{t^4+1}\,dt &= \frac{\sqrt{2}}{2}\left[-\frac{1}{2}\ln(t^2+\sqrt{2}t+1) +\arctan(\sqrt{2}t+1) \right. \\ &\phantom{= \frac{\sqrt{2}}{2}[.} \left.+ \frac{1}{2}\ln(t^2-\sqrt{2}t+1) +\arctan(\sqrt{2}t-1)\right]+C\end{aligned}\]Back substitute \(\large t^2=\tan x \implies t=\sqrt{\tan x}\) to get that
\[\begin{aligned}&\phantom{=}\int\sqrt{\tan x}\,dx\\ &= \frac{\sqrt{2}}{4}\left[-\ln(\tan x+\sqrt{2}\sqrt{\tan x}+1) +2\arctan(\sqrt{2}\sqrt{\tan x}+1) \right. \\ &\phantom{= \frac{\sqrt{2}}{4}[.} \left.+ \ln(\tan x-\sqrt{2}\sqrt{\tan x}+1) +2\arctan(\sqrt{2}\sqrt{\tan x}-1)\right]+C\end{aligned}\]
----------

<sarcasm>That was fun</sarcasm> (and clearly a lot of work). There were a couple points where I skipped some steps (because I was getting tired typing all of this out), so let me know if there are any parts of this you'd like me to explain better.

I hope this makes sense (once you take the time to go over what I did here). :-)

Answer 13

now that is amazing!!

Answer 14

i'm trying to cut and paste this so I can print it out. is there any way of doing this?

Answer 15

If you want, I can take the time to make a pdf version of this (shouldn't take more than 5-10 minutes).

Answer 16

yes, absolutely. this is the best piece of math i've seen on this website

well done. did you use any sources?

Answer 17

Not really; I used Mathematica to check a couple things along the way, but otherwise it was all from my head. XD

Answer 18

thats great. i'll wait for your pdf

Answer 19

Working on it right now. :-)

Answer 20

Enjoy! XD

Answer 21

fantastic. thank you. I'll let you know if i see any errors :)

Answer 22

cool buddy how did you type all these integrals ,did u use any software or did you just typed in each and everything ?
@ChristopherToni ; u are really awesome.

Answer 23

It turns out that Open Study supports the use of \(\large \LaTeX\) through MathJax. For the pdf file I attached afterwords, I have a \(\large \LaTeX\) distribution installed on my PC (called MiKTeX) and an editor. It's all freeware and easy to find via google. :-)

Answer 24

thank you very much. :)

Answer 25

To see the \(\large \LaTeX\) source of each of the math equations in my post (or in anyone else's post), right click the formatted math equation and select the "show math as" dropdown menu and then select the option "TeX commands".

Answer 26

are there any softwares which recognises your writing and convert into text. are you aware of any such softwares @ChristopherToni ?

Answer 27

Hmmm...I'm not sure. :-/

Answer 28

I wonder how could you type all those integrals so easily !!!

Answer 29

6 years worth of experience typesetting math equations with \(\large \LaTeX\) certainly plays a role in that. XD

Answer 30

:)