You're integrating me

$\begin{aligned} I & = \int_0^\frac \pi 2 \sin^{256} x\ dx \\ & = \int_0^\frac \pi 2 \sin^{256} x \cos^0 x\ dx & \small \color{#3D99F6} \text{Beta function B}(m,n) = 2\int_0^\frac \pi 2 \sin^{2m-1}x \cos^{2n-1} x\ dx \\ & = \frac {\text B\left(\frac {257}2, \frac 12\right)}2 & \small \color{#3D99F6} \text{and B}(m,n) = \frac {\Gamma(m)\Gamma(n)}{\Gamma(m+n)} \text{, where }\Gamma (\cdot) \text{ denotes the gamma function.} \\ & = \frac {\Gamma \left(\frac {257}2 \right)\Gamma \left(\frac 12 \right)}{2\Gamma (129)} & \small \color{#3D99F6} \text{Note that }\Gamma \left(\frac 12 \right) = \sqrt \pi, \Gamma (1+x) = x\Gamma (x), \Gamma (n) = (n-1)! \\ & = \frac {\frac {255!!}{128^2} \sqrt \pi \times \sqrt \pi}{2\times 128!} & \small \color{#3D99F6} \text{and that }2^n n! = (2n)!! \\ & = \frac {255!! \pi}{2(256!!)} \end{aligned}$

Therefore, $A+B = 255+256 = \boxed{511}$ .

---

References:

• Beta function
• Gamma function

Recall the reduction formula : $\displaystyle \large \int \sin^{n}{x}dx = -\frac{1}{n}\sin^{n-1}{x}\cos{x}+\frac{n-1}{n}\int\sin^{n-2}{x} dx$

$\large\displaystyle\int_{0}^{\frac{\pi}{2}} \sin^{256}{x} dx = [-\frac{1}{256}\sin^{255}{x}\cos{x}]_{0}^{\frac{\pi}{2}}+\frac{255}{256}\int_{0}^{\frac{\pi}{2}} \sin^{254}{x} dx \\ = 0 + \frac{255}{256} \left( [-\frac{1}{254}\sin^{253}{x}\cos{x}]_{0}^{\frac{\pi}{2}}+\frac{254}{253}\int_{0}^{\frac{\pi}{2}} \sin^{252}{x} dx \right) \\= \frac{255}{256} \left( 0 +\frac{254}{253}\left( [-\frac{1}{252}\sin^{251}{x}\cos{x}]_{0}^{\frac{\pi}{2}}+\frac{252}{251}\int_{0}^{\frac{\pi}{2}} \sin^{250}{x} dx \right)\right) \\= \frac{255}{256}\cdot\frac{253}{254}\cdot\frac{251}{252}\cdots\frac{1}{2}\int_{0}^{\frac{\pi}{2}}\sin^{0}{x} dx \\= \frac{255!!}{256!!}\frac{\pi}{2} \\ \therefore A+B=255+256 = 511$