Fun with 2015 and 2016!!

The substitution $x = -u$ tells us that $I \; = \; \int_{-\frac12\pi}^{\frac12\pi} \frac{1}{2015^x+1} \frac{\sin^{2016}x}{\sin^{2016}x + \cos^{2016}x}\,dx \; = \; \int_{-\frac12\pi}^{\frac12\pi} \frac{2015^u}{2015^u + 1} \frac{\sin^{2016}u}{\sin^{2016}u + \cos^{2016}u}\,du \;,$ so that $I \; = \; \tfrac12 \int_{-\frac12\pi}^{\frac12\pi} \frac{\sin^{2016}x}{\sin^{2016}x + \cos^{2016}x}\,dx \; = \; \int_0^{\frac12\pi} \frac{\sin^{2016}x}{\sin^{2016}x + \cos^{2016}x}\,dx \;.$ The substitution $x = \tfrac12\pi - y$ now gives $I \; = \; \int_0^{\frac12\pi} \frac{\cos^{2016}y}{\sin^{2016}y + \cos^{2016}y}\,dy \;,$ so that $I \; = \; \tfrac12 \int_0^{\frac12\pi} \,dx \; = \; \tfrac14\pi \; = \; \boxed{0.7853981634} \;.$