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Solution

Let $u=\tan x$, then $\sin x=\dfrac{u}{\sqrt{1+u^{2}}}$ and $du=\sec^{2}x\,dx$. $\int_{0}^{1}\dfrac{u}{\sqrt{1+u^{2}}}du=\left[\sqrt{1+u^{2}}\right]_{0}^{1}=\sqrt2-1$. Hence $a=-1$.