Summer of 2019

$\beta(x,y)=\displaystyle\int_{0}^{\frac{\pi}{2}} \sin^{2x-1}t\cos^{2y-1}t \, dt \implies 2x-1=2019, 2y-1=0 \implies x=1010, y=\frac{1}{2}$

$I=\displaystyle\int_{0}^{\frac{\pi}{2}}\sin^{2019}x \, dx = \frac{\beta(1010,\frac{1}{2})}{2}=\frac{\Gamma(1010)\cdot\Gamma(\frac{1}{2})}{2\cdot\Gamma(\frac{2021}{2})}$

We have $\Gamma(s+1)=s\Gamma(s)$ so applying this recurrence $1010$ times means $\Gamma(\frac{2021}{2})=\frac{2019!!}{2^{1010}}\cdot\Gamma(\frac{1}{2})$

$\implies I=\frac{\Gamma(1010)\cdot\Gamma(\frac{1}{2})}{\frac{2019!!}{2^{1009}}\cdot\Gamma(\frac{1}{2})}=\frac{2^{1009}\cdot 1009!}{2019!!} \implies a+b+c+d=\boxed{4039}$

Similar solution with @William Allen 's

$\begin{aligned} I & = \int_0^\frac \pi 2 \sin^{2019} x \ dx \\ & = \int_0^\frac \pi 2 \sin^{2019} x \cos^0 x \ dx & \small \color{#3D99F6} \text{Beta function B}(m,n) = 2 \int_0^\frac \pi 2 \sin^{2m-1} x \cos ^{2n -1} x \ dx \\ & = \frac 12 \text B \left(1010, \frac 12 \right) & \small \color{#3D99F6} \text{and B}(m,n) = \frac {\Gamma (m) \Gamma (n)}{\Gamma(m+n)} \text{, where }\Gamma (\cdot) \text{ is gamma function.} \\ & = \frac {\color{#3D99F6}\Gamma (1010) \Gamma \left(\frac 12 \right)}{2 \color{#D61F06} \Gamma \left(\frac {2021}2 \right)} & \small \color{#3D99F6} \text{Note that } \Gamma (n) = (n-1)!, \Gamma \left(\frac 12 \right) = \sqrt \pi \\ & = \frac {\color{#3D99F6}1009! \sqrt \pi}{2\times \color{#D61F06} \frac {2019!!}{2^{1010}}\sqrt \pi} & \small \color{#3D99F6} \text{and } \Gamma (1+x) = x\Gamma (x) \\ & = \frac {2^{1009}1009!}{2019!!} \end{aligned}$

Therefore, $a+b+c+d = 2+1009+1009+2019 = \boxed{4039}$ .

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References:

• Beta function
• Gamma function