$$\int\sqrt{1-\tan x} \, dx$$
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?
Best Answer
Using your substitution $u^2=1-\tan x$:$$\int \sqrt{1-\tan x}dx=-\int\frac{2u^2}{(u^2-1)^2+1}du=-\int \frac{u^2+\sqrt 2}{u^4-2u^2+2}du-\int \frac{u^2-\sqrt 2}{u^4-2u^2+2}du$$ $$=-\int \frac{1+\frac{\sqrt 2}{u^2}}{\left(u-\frac{\sqrt{2}}{u}\right)^2-2+2\sqrt 2}du-\int \frac{1-\frac{\sqrt 2}{u^2}}{\left(u+\frac{\sqrt{2}}{u}\right)^2-2-2\sqrt 2}du$$ $$=-\frac{1}{\sqrt{2\sqrt 2-2}}\arctan\left(\frac{u-\frac{\sqrt 2}{u}}{\sqrt{2\sqrt 2-2}}\right)-\frac{1}{2\sqrt{2\sqrt 2+2}}\ln\left(\frac{u+\frac{\sqrt{2}}{u}-\sqrt{2+2\sqrt 2}}{{u+\frac{\sqrt{2}}{u}+\sqrt{2+2\sqrt 2}}}\right)+C$$