Trigcky integral

Since @Naren Bhandari has posted part of the solution, here it goes how to prove that

$\prod_{k=1}^{\infty} \cos \left ( \frac{x}{2^k} \right ) = \frac{\sin x}{x}.$

Observe that

$\prod_{k=1}^{N} \cos \left ( \frac{x}{2^k} \right ) = \cos \left ( \frac{x}{2} \right ) \cos \left ( \frac{x}{4} \right ) \cdots \cos\left ( \frac{x}{2^N} \right )$

may be written as

$\frac{\sin x}{2 \sin \left (\frac{x}{2} \right )} \frac{\sin \left ( \frac{x}{2} \right )}{2 \sin \left (\frac{x}{4} \right )} \cdots \frac{\sin \left ( \frac{x}{2^{(N-1)}} \right )}{2 \sin \left (\frac{x}{2^N} \right )}= \frac{\sin x}{2^N \sin\left ( \frac{x}{2^N} \right )},$

since it is well-known that $\sin(2 \alpha) = 2 \sin( \alpha ) \cos( \alpha ).$ Therefore if we take the limit

$\lim_{N \rightarrow \infty} \prod_{k=1}^{N} \cos \left ( \frac{x}{2^k} \right ) = \lim_{N \rightarrow \infty} \frac{\sin x}{2^N \sin\left ( \frac{x}{2^N} \right )} = \sin x \lim_{N \rightarrow \infty} \frac{2^{-N}}{ \sin\left ( \frac{x}{2^N} \right )},$

we may use L'Hospital's rule to get

$\sin x \lim_{N \rightarrow \infty} \frac{2^{-N}}{ \sin\left ( \frac{x}{2^N} \right )} = \sin x \lim_{N \rightarrow \infty} \frac{\frac{\mathrm{d} }{\mathrm{d} N} 2^{-N}}{\frac{\mathrm{d} }{\mathrm{d} N} \sin\left ( \frac{x}{2^N} \right )} = \sin x \lim_{N \rightarrow \infty} \frac{2^{-N} (-N) \ln 2}{x \cos\left ( \frac{x}{2^N} \right ) 2^{-N} (-N) \ln 2} = \frac{\sin x}{x}.$

Since $\dfrac{\sin x}{x} =\lim_{N\to \infty} \prod_{k=1}^{N}\cos \dfrac{x}{2^k} \implies \sin x =\lim_{N\to \infty} \prod_{k=1}^{N}x\cdot \cos \dfrac{x}{2^k}$ and hence $\int_{0}^{\frac{1}{2}\pi}\lim_{N\to \infty} \prod_{k=1}^{N}x\cdot \cos \dfrac{x}{2^k}\,dx = \int_{0}^{\frac{1}{2}\pi} \sin x\,dx = 1$