Với mỗi số $k>0$, đặt...

Cách 1:
Đặt $x=\sqrt{k}\sin t,t\in \left[ -\dfrac{\pi }{2};\dfrac{\pi }{2} \right]\Rightarrow \text{d}x=\sqrt{k}\cos t\text{d}t$
Đổi cận: $x=-\sqrt{k}\Rightarrow t=-\dfrac{\pi }{2},x=\sqrt{k}\Rightarrow t=\dfrac{\pi }{2}$
${{I}_{k}}=\int\limits_{-\dfrac{\pi }{2}}^{\dfrac{\pi }{2}}{\sqrt{k\left( 1-{{\sin }^{2}}t \right)}\sqrt{k}\cos t\text{d}t=}k\int\limits_{-\dfrac{\pi }{2}}^{\dfrac{\pi }{2}}{\left| \cos t \right|\cos t\text{d}t=}k\int\limits_{-\dfrac{\pi }{2}}^{\dfrac{\pi }{2}}{{{\cos }^{2}}t\text{d}t=}\dfrac{k}{2}\int\limits_{-\dfrac{\pi }{2}}^{\dfrac{\pi }{2}}{\left( 1+\cos 2t \right)\text{d}t}$
$=\dfrac{k}{2}\left( t+\dfrac{\sin 2t}{2} \right)\left| \begin{matrix}
\dfrac{\pi }{2} \\
-\dfrac{\pi }{2} \\
\end{matrix}=\dfrac{k\pi }{2} \right.$
Vậy $\sum\limits_{k=1}^{12}{\dfrac{k\pi }{2}=\dfrac{\pi }{2}\left( 1+2+3+...+12 \right)=\dfrac{\pi }{2}\left( \dfrac{12.13}{2} \right)=39\pi }$
Cách 2:
$y=\sqrt{k-{{x}^{2}}}$ là nửa đường tròn phía trên Ox, có bán kính $\sqrt{k}$
${{I}_{1}}+{{I}_{2}}+{{I}_{3}}+...+{{I}_{12}}$ = $\sum\limits_{k=1}^{12}{\dfrac{1}{2}\pi k=\dfrac{1}{2}\pi \left( 1+2+3+...+12 \right)=\dfrac{1}{2}\pi \left( \dfrac{12.13}{2} \right)=39\pi }$