How do I integrate unknown powers?

Let I1 = the given integral and I2=integral where numerator is 2^cosx instead of sin x. I2=I1 because sinx and cosx are identical with just the starting and ending point of limits interchanged. That is between 0 to pi/2, whatever happened to cosine on x will happen to sine on (pi/2 - x) but the area covered will be the same. So, I1=I2=I. I1 + I2 = 2I = integral of 1 dx = pi/2 . So, I = pi/4

use property of definite integral f(x) = f(a+b-x) where a is lower limit and b is upper limit then add both it becomes 2I= integral a to b simple dx ( as denominator remains same while numerator changes to 2 to power cosx as cos(0+pi/2-x) = sinx) answer comes out to be pi/4

Let $\displaystyle\int_{0}^{\frac{\pi}{2}} \frac{2^{sin(x)}}{2^{sin(x)}+2^{cos(x)}}dx=A$ . By definite integral property $\displaystyle\int_{0}^{\frac{\pi}{2}} \frac{2^{sin(x)}}{2^{sin(x)}+2^{cos(x)}}dx=\displaystyle\int_{0}^{\frac{\pi}{2}} \frac{2^{sin(\frac{\pi}{2}-x)}}{2^{sin(\frac{\pi}{2}-x)}+2^{cos(\frac{\pi}{2}-x)}}=\displaystyle\int_{0}^{\frac{\pi}{2}} \frac{2^{cos(x)}}{2^{cos(x)}+2^{sin(x)}}dx=A$ . So $A=\frac{1}{2}(\displaystyle\int_{0}^{\frac{\pi}{2}} \frac{2^{sin(x)}}{2^{sin(x)}+2^{cos(x)}}dx+\displaystyle\int_{0}^{\frac{\pi}{2}} \frac{2^{cos(x)}}{2^{cos(x)}+2^{sin(x)}}dx)=\frac{1}{2}\displaystyle\int_{0}^{\frac{\pi}{2}}dx=\frac{\pi}{4}$