Google
Câu hỏi: Nếu đặt $x=a\sin t$ thì tích phân $\int\limits_{0}^{a}{\dfrac{1}{\sqrt{{{a}^{2}}-{{x}^{2}}}}dx, \left( a>0 \right)}$ trở thành tích phân nào dưới đây?
A. $\int\limits_{0}^{\dfrac{\pi }{2}}{dt}$.
B. $\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{1}{a}dt}$.
C. $\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{a}{t}dt}$.
D. $\int\limits_{0}^{\dfrac{\pi }{4}}{dt}$.
A. $\int\limits_{0}^{\dfrac{\pi }{2}}{dt}$.
B. $\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{1}{a}dt}$.
C. $\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{a}{t}dt}$.
D. $\int\limits_{0}^{\dfrac{\pi }{4}}{dt}$.
Đặt $x=a\sin t\Rightarrow dx=a\cos t.dt$
Ta có $\sqrt{{{a}^{2}}-{{x}^{2}}}=\sqrt{{{a}^{2}}\left( 1-{{\sin }^{2}}t \right)}=\sqrt{{{a}^{2}}c\text{o}{{\text{s}}^{2}}t}=a\left| \cos t \right|$
Đổi cận: $\left\{ \begin{aligned}
& x=0\to t=0 \\
& x=a\to t=\dfrac{\pi }{2} \\
\end{aligned} \right.$
Suy ra $\int\limits_{0}^{a}{\dfrac{1}{\sqrt{{{a}^{2}}-{{x}^{2}}}}dx=}\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{1}{\sqrt{{{a}^{2}}\left( 1-{{\sin }^{2}}t \right)}}}.a\cos t.dt=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{1}{a\left| \cos t \right|}}.a\cos t.dt=\int\limits_{0}^{\dfrac{\pi }{2}}{dt}$.
Ta có $\sqrt{{{a}^{2}}-{{x}^{2}}}=\sqrt{{{a}^{2}}\left( 1-{{\sin }^{2}}t \right)}=\sqrt{{{a}^{2}}c\text{o}{{\text{s}}^{2}}t}=a\left| \cos t \right|$
Đổi cận: $\left\{ \begin{aligned}
& x=0\to t=0 \\
& x=a\to t=\dfrac{\pi }{2} \\
\end{aligned} \right.$
Suy ra $\int\limits_{0}^{a}{\dfrac{1}{\sqrt{{{a}^{2}}-{{x}^{2}}}}dx=}\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{1}{\sqrt{{{a}^{2}}\left( 1-{{\sin }^{2}}t \right)}}}.a\cos t.dt=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{1}{a\left| \cos t \right|}}.a\cos t.dt=\int\limits_{0}^{\dfrac{\pi }{2}}{dt}$.
Đáp án A.