This can be done with a standard $\sin$ substitution: $(1)\ x=a\sin(\theta)$ followed by a change of variables: $(2)\ \phi=2\theta$. $ \begin{align} \int_{-a}^a2\sqrt{a^2-x^2}\,\mathrm{d}x &=\int_{-\pi/2}^{\pi/2}2a\cos(\theta)\,\mathrm{d}a\sin(\theta)\tag{1}\\ &=2a^2\int_{-\pi/2}^{\pi/2}\cos^2(\theta)\,\mathrm{d}\theta\\ &=2a^2\int_{-\pi/2}^{\pi/2}\frac{1+\cos(2\theta)}{2}\,\mathrm{d}\theta\\ &=2a^2\int_{-\pi}^\pi\frac{1+\cos(\phi)}{4}\,\mathrm{d}\phi\tag{2}\\ &=2a^2\left[\frac{\phi+\sin(\phi)}{4}\right]_{\phi=-\pi}^{\phi=\pi}\\ &=\pi a^2 \end{align} $ However, the answer you show in the question looks like the answer that comes from an integration by parts: $u=\sqrt{a^2-x^2}$ and $\mathrm{d}v=\mathrm{d}x$ so that $v=x$ and $\mathrm{d}u=-\frac{x\,\mathrm{d}x}{\sqrt{a^2-x^2}}$ $ \begin{align} \int2\sqrt{a^2-x^2}\,\mathrm{d}x &=2x\sqrt{a^2-x^2}+\int\frac{2x^2}{\sqrt{a^2-x^2}}\,\mathrm{d}x\\ &=2x\sqrt{a^2-x^2}-\color{#C00000}{\int\frac{2(a^2-x^2)}{\sqrt{a^2-x^2}}\,\mathrm{d}x}+\int\frac{2a^2}{\sqrt{a^2-x^2}}\,\mathrm{d}x\tag{3} \end{align} $ Adding the integral in red to both sides of $(3)$ and dividing by $2$ yields $ \begin{align} \int2\sqrt{a^2-x^2}\,\mathrm{d}x &=\frac12\left[2x\sqrt{a^2-x^2}+\int\frac{2a^2}{\sqrt{a^2-x^2}}\,\mathrm{d}x\right]\\ &=\frac12\left[2x\sqrt{a^2-x^2}+2a^2\,\sin^{-1}(x/a)\right]+C\tag{4} \end{align} $ Evaluating $(4)$ at the limits of integration yields $ \begin{align} \int_{-a}^a2\sqrt{a^2-x^2}\,\mathrm{d}x &=\left[x\sqrt{a^2-x^2}+a^2\,\sin^{-1}(x/a)\right]_{-a}^a\\ &=\pi a^2 \end{align} $ Not quite what you got, but of the same form.