Đáp án:
\[I = 16\int\limits_0^{\frac{\pi }{2}} {{{\sin }^2}tdt} \]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
I = \int\limits_0^4 {\sqrt {16 - {x^2}} dx} \\
x = 4\cos t \Rightarrow \left\{ \begin{array}{l}
dx = 4.\left( {{\mathop{\rm cost}\nolimits} } \right)'dt = - 4\sin tdt\\
x = 0 \Rightarrow t = \frac{\pi }{2}\\
x = 4 \Rightarrow t = 0
\end{array} \right.\\
\Rightarrow I = \int\limits_{\frac{\pi }{2}}^0 {\sqrt {16 - {{\left( {4\cos t} \right)}^2}} .\left( { - 4\sin tdt} \right)} \\
= - \int\limits_0^{\frac{\pi }{2}} {4.\sqrt {1 - {{\cos }^2}t} .\left( { - 4\sin tdt} \right)} \\
= \int\limits_0^{\frac{\pi }{2}} {4.\sin t.4\sin tdt} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {t \in \left[ {0;\frac{\pi }{2}} \right] \Rightarrow \sin t > 0} \right)\\
= 16\int\limits_0^{\frac{\pi }{2}} {{{\sin }^2}tdt}
\end{array}\)
