Connecting series and integrals

The substitution $u \; = \; \sqrt{\frac{1-\sin x}{1+\sin x}}$ makes $\int_0^{\frac12\pi} \tan^{-1}\sqrt{\frac{\sin x}{1+\sin x}}\,dx \; = \; 2 \int_0^1 \tan^{-1}\sqrt{\tfrac{1-u^2}{2}}\,\frac{du}{1+u^2}$ and this second integral is evaluated here , so that $\int_0^{\frac12\pi} \tan^{-1}\sqrt{\frac{\sin x}{1+\sin x}}\,dx \; = \; 2 \times \frac{\pi^2}{24} \; = \; \frac{\pi^2}{12} \; = \; \tfrac12\zeta(2)$ making the answer $2+2=\boxed{4}$ .