Sines are Cosines

Let $I$ denote the integral. $\begin{aligned} I = & \int_0^{\frac{\pi}{2}} \frac{\sin^n(x)}{\sin^n(x)+\cos^n(x)}\,dx \\ I= & \int_0^{\frac{\pi}{2}} \frac{\sin^n(\frac{\pi}{2} - x)}{\sin^n(\frac{\pi}{2} -x) + \cos^n(\frac{\pi}{2} - x)} \,dx \text{ via } \color{#3D99F6} \int_a^bf(x) \,dx = \int_a^bf(a+b-x)\,dx\\ I = &\int_0^{\frac{\pi}{2}} \frac{\cos^n(x)}{\sin^n(x)+\cos^n(x)}\,dx \\ 2I = & \int_0^{\frac{\pi}{2}} \frac{\sin^n(x) + \cos^n(x)}{\sin^n(x)+\cos^n(x)}\,dx \\ 2I = & \frac{\pi}{2}\\ I = & \frac{\pi}{4} \\ &\boxed{\frac{a}{b} = 1} \end{aligned}$

Let $n = 0$ , then

$\int_0^{\frac{\pi}{2}} \frac{\sin^0 (x)}{\sin^0(x) + cos^0(x)} = \int_0^{\frac{\pi}{2}} \frac{1}{2} = \frac{\pi}{4}$

$\displaystyle a = b \ne 0, \text { so } \frac { \pi ^ { a } } { 4 ^ { b } } \rightarrow \frac { a } { b } = \boxed { 1 }$ .