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$$\int\sqrt{1-\tan x} \, dx$$ is a very interesting integral. I attempted to evaluate it with the substitution $u^2=1-\tan x$ and then obtaining partial fractions. However, the coefficients are extremely complicated.

Is there an easier way to do it?

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  • $\begingroup$ What do you mean by evaluating without paper? Like I must do it in my head? $\endgroup$ – Nyssa May 17 at 19:31
  • $\begingroup$ @Threesidedcoin without using a lot of paper $\endgroup$ – Jethro May 17 at 22:15
  • $\begingroup$ @user90369 Why not put it back on now? It will probably not be downvoted again. $\endgroup$ – Jethro May 17 at 22:16
  • $\begingroup$ It's o.k. now, there are other answers, that's enough. ;) $\endgroup$ – user90369 May 18 at 5:10
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    $\begingroup$ It's only $\displaystyle\int\sqrt{1-\tan x}~dx = i \left( \sqrt{1+i}~\tanh^{-1}\frac{u}{\sqrt{1+i}} - \sqrt{1-i}~\tanh^{-1}\frac{u}{\sqrt{1-i}} \right) + C~$, it's easy to get it. But if you look at the solution of Three sided coin it's also a very short method, and better because it's without $~i~$. I think that's the best! ;) The coefficients are as they are, you always have to solve the equation $~z^2-2z+2=0~$ . $\endgroup$ – user90369 May 18 at 18:37
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Using your substitution $u^2=1-\tan x$:$$\int \sqrt{1-\tan x}dx=-\int\frac{2u^2}{(u^2-1)^2+1}du=-\int \frac{u^2+\sqrt 2}{u^4-2u^2+2}du-\int \frac{u^2-\sqrt 2}{u^4-2u^2+2}du$$ $$=-\int \frac{1+\frac{\sqrt 2}{u^2}}{\left(u-\frac{\sqrt{2}}{u}\right)^2-2+2\sqrt 2}du-\int \frac{1-\frac{\sqrt 2}{u^2}}{\left(u+\frac{\sqrt{2}}{u}\right)^2-2-2\sqrt 2}du$$ $$=-\frac{1}{\sqrt{2\sqrt 2-2}}\arctan\left(\frac{u-\frac{\sqrt 2}{u}}{\sqrt{2\sqrt 2-2}}\right)-\frac{1}{2\sqrt{2\sqrt 2+2}}\ln\left(\frac{u+\frac{\sqrt{2}}{u}-\sqrt{2+2\sqrt 2}}{{u+\frac{\sqrt{2}}{u}+\sqrt{2+2\sqrt 2}}}\right)+C$$

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  • $\begingroup$ So, how is this different from what the OP attempted? $\endgroup$ – Oscar Lanzi May 18 at 9:23
  • $\begingroup$ @OscarLanzi who said it should be different? The requirement was to not use up paper. $\endgroup$ – Nyssa May 18 at 9:28

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