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Just some thinking in the shower this morning (I'm a little rusty so forgive me for any errors!). We have the Riemann-Liouville integral: D_{t}^{-q} f(t) = \frac{1}{\Gamma (q)} \int_{0}^{t} (t - \tau)^{q - 1} f(\tau) d \tau So if I take the Laplace transform we get: \mathcal{L} \{...

\text{By writing }(cosx)^{2n}=(1-sin^2x)^n\text{, show that }\int_{0}^{\frac{\pi}{2}}(cosx)^{2n+1}dx=\sum_{k=0}^{n}\frac{(-1)^k}{2k+1}\binom{n}{k} I can see why it would equal that, but I just can't get there >.< Probably something obvious but I'm just not seeing it.