Evaluate $\int_0^1 \frac {1- \tan(x)}{\sqrt{x}+\tan(x)}dx$ I am having an extremely hard time finding the anti-derivative of this function:
$$f(x)=\dfrac{1-\tan(x)}{\sqrt{x}+\tan(x)}$$
while the lower bound is 0 and the upper bound is 1.
I tried to expand by $\cos(x)$ so it would look like
$$f(x)=\dfrac{\cos(x)-\sin(x)}{\sqrt{x}\cos(x)+\sin(x)}$$
and then linearly separate it into 2 integrals but it didn't help much at all. Any help would be appreciated, thank you very much.
 A: This is not an answer but it is too long for a comment.
If you had "extremely hard time finding the anti-derivative of this function", now we are two !
I do not think that we could find the antiderivative 
$$I=\int\frac {1- \tan(x)}{\sqrt{x}+\tan(x)}\,dx$$ even using special functions.
What I first did is to let $x=y^2$ to work
$$J=\int\frac{2 y \left(1-\tan \left(y^2\right)\right)}{y+\tan \left(y^2\right)}\,dy$$ If you plot the integrand for $0\leq y\leq 1$, you could notice that it is "close" to $2-2y$ and then, over this range, the definite integral should be "close" to $1$ (smaller that $1$ because or the curvature of the integrand).
What I tried was to approximate the integrand looking for $[n,m]$ Padé approximants (built around $y=0$) which could be workable for integration. However, the problem I faced is that, until $m=4$, I just obtain the linear approximation mentioned above. For $n=2$, the denominator cancels in the range. So, I kept the simplest $(n=3,m=4)$ which gives, as an approximation,
$$\frac{2 y \left(1-\tan \left(y^2\right)\right)}{y+\tan \left(y^2\right)}=\frac {2-2 y^2-\frac{2 }{3}y^3 } {1+y-\frac{1}{3}y^3-\frac{1}{3}y^4 }=6\frac {1- y^2-\frac{1 }{3}y^3 } {(1+y)(3-y^3)}$$ The last expression can be decomposed using partial fraction since,$a,b,c$ being the roots of $y^3=3$, 
$$\frac{1- y^2-\frac{1 }{3}y^3 }{(y+1) (y-a) (y-b) (y-c)}=\frac{-a^3-3 a^2+3}{3 (a+1) (a-b) (a-c) (y-a)}+\frac{-b^3-3 b^2+3}{3 (b+1) (b-a)
   (b-c) (y-b)}+\frac{-c^3-3 c^2+3}{3 (c+1) (c-a) (c-b) (y-c)}-\frac{1}{3 (a+1)
   (b+1) (c+1) (y+1)}$$ Integrating and recombining everything leads to the "small monster"
$$\frac{1}{12} \left(3 \left(\log \left(\frac{16}{9}\right)+\sqrt[3]{3}
   \left(\sqrt[3]{3}-1\right) \log \left(\frac{1}{2} \left(2+3 \sqrt[3]{3}-3\
   3^{2/3}\right)\right)+2 \sqrt[6]{3} \left(3+3^{2/3}\right) \tan
   ^{-1}\left(\frac{2+\sqrt[3]{3}}{3^{5/6}}\right)\right)-\sqrt[6]{3}
   \left(3+3^{2/3}\right) \pi \right)\approx 0.922454$$ while a numerical integration would lead to $\approx {0.929600}$.
Not very fantastic, I agree.
Edit
Just for the fun of it, I built the $[2,2]$ Padé approximant around $y=\sqrt{\frac{\pi }{4}}$ (I shall not type the the value of the coefficients). So, $J$ can be computed and the result is $\approx 0.928856$ (much better than the previous one).
