Summing up factorials (3)

Multiply $\sin(x) = \sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)!}x^{2k+1}, \qquad \cos(x) = \sum_{i=0}^\infty \frac{(-1)^i}{(2i)!}x^{2i}$ to find $\sin(x)\cos(x) = \sum_{k=0}^\infty\sum_{i=0}^\infty \frac{(-1)^{i+k}}{(2k+1)!(2i)!}x^{2i+2k+1}.$

Now, integrate both sides from $x=0$ to $x=1$ : $\int_0^1 \sin(x)\cos(x)\,dx = \int_0^1 \left(\sum_{k=0}^\infty\sum_{i=0}^\infty \frac{(-1)^{i+k}}{(2k+1)!(2i)!}x^{2i+2k+1}\right)\,dx = \sum_{k=0}^\infty\sum_{i=0}^\infty \frac{(-1)^{i+k}}{(2k+1)!(2i)!}\int_0^1 x^{2i+2k+1}\,dx$ $\frac{1}{2}\sin^2(1) = \sum_{k=0}^\infty\sum_{i=0}^\infty \frac{(-1)^{i+k}}{(2k+1)!(2i)!(2i+2k+2)}$

Finally, just multiply both sides by $2$ : $\sin^2(1) = \sum_{k=0}^\infty\sum_{i=0}^\infty \frac{(-1)^{i+k}}{(2k+1)!(2i)!(i+k+1)}$

Therefore, $a=1, b=2$ , so $\boxed{a+b=3}$ .