<s>[INST] Find the value of $\int_\frac{-\pi}{2}^\frac{\pi}{2} 2^{\sin x} \text dx +\int_\frac{5}{2}^4 \sin^{-1}(\log_2(x-2)) \text dx$ 
Find the value of $$\int_\frac{-\pi}{2}^\frac{\pi}{2}2^{\sin x} \text dx +\int_\frac{5}{2}^4 \sin^{-1}(\log_2(x-2)) \text dx.$$

I tried to solve it using the concept: $$\int_{a}^b f(x)\text dx=\int_{a}^b f(a+b-x)\text dx.$$ 
My step is as follow:
$$I=\int_\frac{-\pi}{2}^\frac{\pi}{2}2^{\sin x}dx=\int_\frac{-\pi}{2}^\frac{\pi}{2}2^{-\sin x}dx.$$
$$I=\int_\frac{-\pi}{2}^\frac{\pi}{2} f(x)dx=\int_\frac{-\pi}{2}^\frac{\pi}{2}f(-x)dx$$
 [/INST] It seems this problem is designed to use the formula
$$\int_a^bf(x)\;dx + \int_{f(a)}^{f(b)}f^{-1}(x)\;dx = bf(b)-af(a)$$
(See here)
You only need to rewrite the second integral using $x\mapsto x-2$ as
$$\int_\frac{5}{2}^4 \sin^{-1}(\log_2(x-2)) \;dx= \int_{\frac 12}^{2}\sin^{-1}(\log_2(x))\;dx$$
So, you get 
$$\int_\frac{-\pi}{2}^\frac{\pi}{2}2^{\sin x} \text dx +\int_\frac{5}{2}^4 \sin^{-1}(\log_2(x-2)) \text dx$$
$$=\int_\frac{-\pi}{2}^\frac{\pi}{2}2^{\sin x} \text dx +  \int_{\frac 12}^{2}\sin^{-1}(\log_2(x))\;dx= \frac{\pi}{2}\cdot 2+ \frac{\pi}{2}\cdot\frac 12 = \frac 54\pi$$
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