Question

A walkthrough of the integration of \(\sqrt{\tan x}\).

Answer 1

\[\int\sqrt{\tan x}dx \]
Start with the substitution method.

\[
u=\sqrt{\tan x}, x=\arctan u^2, dx=\frac{2u}{u^4+1}du\\
\int u\cdot\frac{2u}{u^4+1}du=\int\frac{2u^2}{u^4+1}du\\
\]
Next, use partial fractions.
\[
(u^4+1)=(u^2+\sqrt2u+1)(u^2-\sqrt2u+1)\\
\frac{2u^2}{u^4+1}=\frac{Au+B}{u^2+\sqrt2u+1}+\frac{Cu+D}{u^2-\sqrt2u+1}\\
2u^2=(Au+B)(u^2-\sqrt2u+1)+(Cu+D)(u^2+\sqrt2u+1)\\
2u^2=(A+C)u^3+(B+D-\sqrt2A+\sqrt2C)u^2+(A+C-\sqrt2B+\sqrt2D)u+(B+D)
\]
\[\begin{align}
A+C&=0\\
B+D-\sqrt2A+\sqrt2C&=2\\
A+C-\sqrt2B+\sqrt2D&=0\\
B+D&=0
\end{align}\]
\[C=\frac{\sqrt2}{2}, A=-\frac{\sqrt2}{2}\]
\[\int\left[
\frac{\frac{-\sqrt2}{2}u}{u^2+\sqrt2u+1}+
\frac{\frac{\sqrt2}{2}u}{u^2-\sqrt2u+1}
\right]du\\
\frac{-\sqrt2}{4}\int\frac{2u}{u^2+\sqrt2u+1}du+\frac{\sqrt2}{4}\int\frac{2u}{u^2-\sqrt2u+1}du\\
\frac{-\sqrt2}{4}\int\frac{2u+\sqrt2-\sqrt2}{u^2+\sqrt2u+1}du+\frac{\sqrt2}{4}\int\frac{2u+\sqrt2-\sqrt2}{u^2-\sqrt2u+1}du\\
\frac{-\sqrt2}{4}\left[ \log(u^2+\sqrt2u+1)-\sqrt2\int\frac{1}{u^2+\sqrt2u+1}du\right]\\\quad+\frac{\sqrt2}{4}\left[ \log(u^2-\sqrt2u+1)+\sqrt2\int\frac{1}{u^2-\sqrt2u+1}du\right]
\]
Now, complete the square.
\[
\begin{align}
u^2\pm\sqrt2u+1&=\left(u\pm\frac{\sqrt2}{2}\right)^2+\frac{1}{2}\\
&=\frac{1}{2}\left[\left( \frac{u\pm\frac{\sqrt2}{2}}{\sqrt{\frac{1}{2}}}\right)^2+1
\right]\\
&=\frac{1}{2}\left[\left( \sqrt2u\pm1\right)^2+1
\right]
\end{align}
\]
Rewrite the integrals.
\[
\int\frac{1}{u^2+\sqrt2u+1}du=2\int\frac{1}{\left( \sqrt2u+1\right)^2+1}du\\
\int\frac{1}{u^2-\sqrt2u+1}du=2\int\frac{1}{\left( \sqrt2u-1\right)^2+1}du
\]
Use the substitution method.
\[
t=\sqrt2u+1, u=\frac{\sqrt2}{2}t-\frac{\sqrt2}{2}, du=\frac{\sqrt2}{2}dt\\
2\int\frac{1}{\left( \sqrt2u+1\right)^2+1}=\sqrt2\int\frac{1}{t^2+1}dt\\=\sqrt2\arctan t=\sqrt2\arctan (\sqrt2u+1)\]
\[
t=\sqrt2u-1, u=\frac{\sqrt2}{2}t+\frac{\sqrt2}{2}, du=\frac{\sqrt2}{2}dt\\
2\int\frac{1}{\left( \sqrt2u-1\right)^2+1}=\sqrt2\int\frac{1}{t^2+1}dt\\=\sqrt2\arctan t=\sqrt2\arctan (\sqrt2u-1)
\]
Now simplify.
\[
\frac{-\sqrt2}{4}\left[ \log(u^2+\sqrt2u+1)-\sqrt2\left(\sqrt2\arctan(\sqrt2u+1)\right)\right]\\ \quad +\frac{\sqrt2}{4}\left[ \log(u^2-\sqrt2u+1)+\sqrt2\left(\sqrt2\arctan(\sqrt2u-1)\right)\right]\\

\frac{-\sqrt2}{4}\log(u^2+\sqrt2u+1)+\frac{\sqrt2}{4}\log(u^2-\sqrt2u+1)\\\quad+\frac{\sqrt2}{2}\arctan(\sqrt2u+1)+\frac{\sqrt2}{2}\arctan(\sqrt2u-1)
\]
Put it back in terms of x.
\[
\frac{-\sqrt2}{4}\log(\tan x+\sqrt{2\tan x}+1)+\frac{\sqrt2}{4}\log(\tan x-\sqrt{2\tan x}+1)\\+\frac{\sqrt2}{2}\arctan(\sqrt{2\tan x}+1)+\frac{\sqrt2}{2}\arctan(\sqrt{2\tan x}-1)
\]
And there you have it.

Answer 2

@inkyvoyd

Answer 3

@nb, this is one of the longest integrals and I am glad you had the patience to write so well. It should surely help others :)

Answer 4

You made a mistake!

Answer 5

@inkyvoyd , would you mind elaborating where exactly ?

Answer 6

Yes. The whole fun of this is finding the mistake.

Answer 7

@inkyvoyd, the official OS troll :P

Answer 8

Good, @shivam_bhalla , you figured out what I am doing for the most part.