Are you going to use formula?

The integration $\small\int_{0}^{\frac{\pi}{\sqrt[10]{1024}}} \sin^{\sqrt{64}}x\,dx$ can be simplified to $\small\int_{0}^{\frac{\pi}{2}} \sin^8 x\,dx$ . Before to integrate let us generalize the integration of $\small\int_{0}^{\frac{\pi}{2}} \sin^n x\,dx$ for any integer. $n\geq 1$ .

$\small\int_{0}^{\frac{\pi}{2}} \sin^n x\,dx = \int_{0}^{\frac{\pi}{2}}\sin^{n}. \sin^{n-1} x\,dx \\\int_{0}^{\frac{\pi}{2}} \sin^n x\,dx = - \text{cosx} . \sin^{n-1}x\,dx|_{0}^{\frac{\pi}{2}} +(n-1)\int_{0}^{\frac{\pi}{2}} \cos^2x .\sin^{n-2}x\,dx \phantom{ccccc}{\color{#3D99F6}\text{Integration by parts}} \\ \int_{0}^{\frac{\pi}{2}} \sin^n x\,dx = 0 +(n-1)\int_{0}^{\frac{\pi}{2}} \cos^2x .\sin^{n-2}x\,dx \\\int_{0}^{\frac{\pi}{2}} \sin^n x\,dx =(n-1) \int_{0}^{\frac{\pi}{2}} \sin^{n-2}x\,dx- (n-1)\int_{0}^{\frac{\pi}{2}}\sin^n x\,dx \\ \int_{0}^{\frac{\pi}{2}} \sin^n x\,dx = \frac{n-1}{n}\int_{0}^{\frac{\pi}{2}} \sin^{n-2}x\,dx \\ \implies I = \frac{n-1}{n} I_{n-2} =\frac{n-2}{n}.\frac{n-3}{n-1} I_{n-4} = \frac{n-1}{n}.\frac{n-3}{n-2}.\frac{n-5}{n-4}I_{n-6} = \cdots$ Now integrating continuesly using Integration by parts till $\sin^{\circ} x$ if $n$ is an even number and $\text{sinx}\,dx$ if $n$ is odd number we obtained as $\int_{0}^{\frac{\pi}{2}} \sin^n\,dx = \begin{cases}\frac{n-1}{n}.\frac{n-3}{n-2}.\frac{n-5}{n-4} \cdots \frac{\pi}{2} \phantom{cccc}{\color{#3D99F6}\text{if n is even}} \\ \frac{n-1}{n}.\frac{n-3}{n-2}.\frac{n-5}{n-4} \cdots 1 \phantom{cccc}{\color{#3D99F6}\text{if n > 1 is odd }} \end{cases}$ Since in the case we find $n=8$ an even number so $\int_{0}^{\frac{\pi}{2}} \sin^8x\,dx = \frac{7.5.3.1\pi}{2.4.6.8} =\frac{105}{32(4!)}$ . Hence the value of $a+b+c = 105+32+4=\boxed{141}$ .