-1
$\begingroup$

What is wrong with this solution? https://www.mathcha.io/editor/egPXHEXSj3CD1ip1

$$I=\int \sqrt{\tan x}dx$$ $u=\sqrt{\tan x}\Rightarrow \frac{2u}{u^2+1}du=dx$: $$I=2\int\frac{u^2}{u^2+1}du$$ $$I=2\int\frac{u^2+1-1}{u^2+1}du$$ $$I=2\bigg(\int 1du-\int\frac1{u^2+1}du\bigg)$$ $$I=2\big(u-\arctan u\big)+C$$ $$I=2\sqrt{\tan x}-2\arctan\sqrt{\tan x}+C$$

Thanks!

Please do not link other solutions of this integral, I just want to know why the one I did is incorrect, thanks!

$\endgroup$
  • 1
    $\begingroup$ Figuring out LaTeX is pretty much a requirement for using this site. Plus, you'll gain a great skill if you ever have to write-up math in the future. $\endgroup$ – JonathanZ Nov 24 '18 at 1:09
  • 1
    $\begingroup$ I added the LaTex as an edit. Take a look at it so you can see how it's done. $\endgroup$ – clathratus Nov 24 '18 at 1:16
4
$\begingroup$

$\cos^2 x \neq \frac{1}{1+(\sqrt{\tan{x}})^2}$. $\cos^2 x = \frac{1}{1+(\sqrt{\tan{x}})^4} = \frac{1}{1+u^4}$.

$\endgroup$
0
$\begingroup$

$$u=\sqrt{\tan x}\implies x= \tan ^{-1}\left(u^2\right)\implies dx=\frac{2 u}{u^4+1}\,du$$ $$I=\int \sqrt{\tan x}\,dx=2\int \frac{ u^2}{u^4+1}\,du=\int \frac{ u^2}{\left(u^2-\sqrt{2} u+1\right) \left(u^2+\sqrt{2} u+1\right)}\,du$$ Now, partial fraction decomposition $$I=\frac 1 {\sqrt 2}\left(\int \frac{u}{u^2-\sqrt{2} u+1} \,du-\int \frac{u}{u^2+\sqrt{2} u+1}\,du \right)$$ which becomes to be quite easy.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.