Question

$\underset{0}{\overset{a}{\int }}\sqrt{a^2-x^2}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I={\int }_0^a\sqrt{a^2-x^2}\mathit{d}x\mathit{.} \\ \mathrm{Let}x=a\sin t.\mathrm{Then},dx=a\cos tdt \\ \mathrm{When}x=0,t=0\mathrm{and}x=a,t=\frac{\mathrm{\pi }}{2} \\ \therefore I={\int }_0^a\sqrt{a^2-x^2}\mathit{d}x \\ \Rightarrow I={\int }_0^{\frac{\mathrm{\pi }}{2}}\sqrt{\left(a^2-a^2{\sin }^{2}t\right)}a\cos t\mathit{d}t \\ \Rightarrow I={\int }_0^{\frac{\mathrm{\pi }}{2}}{a}^2{\cos }^{2}tdt \\ \Rightarrow I=a^2{\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{1+\cos 2t}{2}dt \\ \Rightarrow I=\frac{a^2}{2}{\left[t+\frac{\sin 2t}{2}\right]}_0^{\frac{\mathrm{\pi }}{2}} \\ \Rightarrow I=\frac{a^2}{2}\left(\frac{\mathrm{\pi }}{2}-0\right) \\ \Rightarrow I=\frac{{\mathrm{\pi a}}^2}{4} \end{gathered}$$