Integration problem

Suppose that $a,b$ are real with $b > a^2 > 1$ , and consider the integrals $\begin{array}{rcl} \displaystyle I_1(a,b) & = & \displaystyle \int_1^\infty \frac{du}{u^4 + 2(b-2a^2)u^2 + b^2} \; = \; \int_1^\infty \frac{du}{(u^2 + 2au + b)(u^2 - 2au + b)} \\ \displaystyle I_2(a,b) & = & \displaystyle \int_1^\infty \frac{du}{(u^4 + 2(b-2a^2)u^2 + b^2)^2} \; = \; \int_1^\infty \frac{du}{(u^2 + 2au + b)^2(u^2 - 2au + b)^2} \\ \displaystyle I_3(a,b) & = & \displaystyle \int_1^\infty \left(\frac{1}{(u^2 + 2au + b)^2} + \frac{1}{(u^2 - 2au + b)^2}\right)\,du \end{array}$ Now, using partial fractions and standard integrals $\begin{array}{rcl} \displaystyle I_1(a,b) & = & \displaystyle \frac{1}{4ab}\int_1^\infty \left(\frac{u+2a}{u^2 + 2au + b} - \frac{u-2a}{u^2 - 2au + b}\right)\,du \\ & = & \displaystyle \frac{1}{4ab}\int_1^\infty \left(\frac{u+a}{u^2 + 2au + b} - \frac{u-a}{u^2 - 2au + b} + \frac{a}{u^2 + 2au + b} + \frac{a}{u^2 - 2au + b}\right)\,du \\ & = & \displaystyle \frac{1}{4ab}\left[ \tfrac12\ln\left(\frac{u^2 + 2au + b}{u^2 - 2au + b}\right) + \frac{a}{\sqrt{b-a^2}}\tan^{-1}\left(\frac{u+a}{\sqrt{b-a^2}}\right) + \frac{a}{\sqrt{b-a^2}}\tan^{-1}\left(\frac{u-a}{\sqrt{b-a^2}}\right)\right]_1^\infty \\ & = & \displaystyle \frac{1}{4ab}\left[\frac{a \pi}{\sqrt{b-a^2}} - \tfrac12\ln\left(\frac{1+2a+b}{1-2a+b}\right) - \frac{a}{\sqrt{b-a^2}}\tan^{-1}\left(\frac{1+a}{\sqrt{b-a^2}}\right) - \frac{a}{\sqrt{b-a^2}}\tan^{-1}\left(\frac{1-a}{\sqrt{b-a^2}}\right)\right] \\ & = & \displaystyle \frac{1}{4ab}\left[ \frac{a\pi}{\sqrt{b-a^2}} - \tfrac12\ln\left(\frac{1+2a+b}{1-2a+b}\right) - \frac{a}{\sqrt{b-a^2}}\tan^{-1}\left(\frac{2\sqrt{b-a^2}}{b-1}\right)\right] \end{array}$ Similarly we have $\begin{array}{rcl} \displaystyle \frac{1}{(u^4 + 2(b-2a^2)u^2 + b^2)^2} & = & \displaystyle \frac{4a^2+b}{32a^3b^3}\left[ \frac{u+2a}{u^2 + 2au + b} - \frac{u-2a}{u^2 - 2au + b}\right] \\ & & \displaystyle + \frac{1}{16a^2b^2}\left[\frac{2au + (4a^2 - b)}{(u^2 + 2au + b)^2} - \frac{2au - (4a^2 - b)}{(u^2 - 2au + b)^2}\right] \\ & = & \displaystyle \frac{4a^2 + b}{8a^2b^2} \frac{1}{u^4 + 2(b-2a^2)u^2 + b^2} + \frac{1}{16a^2b^2}\left[\frac{2au + (4a^2 - b)}{(u^2 + 2au + b)^2} - \frac{2au - (4a^2 - b)}{(u^2 - 2au + b)^2}\right] \end{array}$ and so $\begin{array}{rcl} \displaystyle I_2(a,b) & = & \displaystyle \frac{4a^2 + b}{8a^2b^2}I_1(a,b) + \frac{1}{16a^2b^2}\int_1^\infty \left\{ 2a\left(\frac{u+a}{(u^2 + 2au + b)^2} - \frac{u-a}{(u^2 - 2au + b)^2}\right) + (2a^2 - b)\left(\frac{1}{(u^2 + 2au + b)^2} + \frac{1}{(u^2 - 2au + b)^2}\right)\right\}\,du \\ & = & \displaystyle \frac{4a^2 + b}{8a^2b^2}I_1(a,b) + \frac{1}{16ab^2}\left[ \frac{1}{u^2 - 2au + b} - \frac{1}{u^2 + 2au + b}\right]_1^\infty + \frac{2a^2 - b}{16a^2 b^2}\int_1^\infty \left(\frac{1}{(u^2 + 2au + b)^2} + \frac{1}{(u^2 - 2au + b)^2}\right)\,du \\ & = & \displaystyle \frac{4a^2 + b}{8a^2b^2}I_1(a,b) - \frac{1}{4b^2\big((1+b)^2 - 4a^2\big)} + \frac{2a^2 - b}{16a^2 b^2}I_3(a,b) \end{array}$ Now, the substitution $u + a = \sqrt{b-a^2}\tan\phi$ shows us that $\begin{array}{rcl} \displaystyle \int_1^\infty \frac{1}{(u^2+ 2au + b)^2}\,du & = & \displaystyle \frac{1}{(b-a^2)^{\frac32}}\int_{\tan^{-1}\frac{a+1}{\sqrt{b-a^2}}}^{\frac12\pi} \cos^2\phi\,d\phi \\ & = & \displaystyle \frac{1}{2(b - a^2)^{\frac32}}\left[\tfrac12\pi - \frac{(1+a)\sqrt{b-a^2}}{1+2a+b} -\tan^{-1}\left(\frac{1+a}{\sqrt{b-a^2}}\right)\right] \end{array}$ so that $I_3(a,b) \; = \; \frac{\pi}{2(b - a^2)^{\frac32}} - \frac{1+b - 2a^2}{(b-a^2)\big((1+b)^2 - 4a^2\big)} - \frac{1}{2(b-a^2)^{\frac32}}\tan^{-1}\left(\frac{2\sqrt{b-a^2}}{b-1}\right)$ Thus we have concrete, if complicated, expressions for $I_1(a,b)$ , $I_2(a,b)$ and $I_3(a,b)$ .

Going back to the original problem, the substitution $u^2 = 1 + \tan x$ gives us the identity $\begin{array}{rcl} \displaystyle \int_0^{\frac12\pi} \frac{\sin^2x}{\sqrt{1 +\tan x}}\,dx & = & \displaystyle 2\int_1^\infty \frac{(u^2-1)^2}{\big((u^2-1)^2+1\big)^2}\,du \\ & = & \displaystyle 2\int_1^\infty \frac{1}{u^4 - 2u^2 + 2}\,du - 2\int_1^\infty \frac{1}{(u^4 - 2u^2 + 2)^2}\,du \\ & = & \displaystyle 2I_1(a,b) - 2I_2(a,b) \end{array}$ where $a \; = \; \sqrt{\frac{1+\sqrt{2}}{2}} \hspace{2cm} b \; = \; \sqrt{2}$ Substituting these values of $a$ and $b$ into the above formulae gives the answer $\boxed{0.418783}$ . Although the above formulae give an exact expression for this integral, it does not seem to have any particularly simple form, and so I have omitted it.