CÂU HỎI: Cho nguyên hàm \( I={\smallint \sqrt {1 – {x^2}} {\mkern 1mu} {\rm{d}}x}\), x thuộc \( \left[ {0;\frac{\pi }{2}} \right]\)  , nếu đặt x = sin t thì nguyên hàm I tính theo biến t trở thành:

Đặt 

\(\begin{array}{l}
x = \sin t \Leftrightarrow dx = \cos t{\mkern 1mu} dt\\
1 – {x^2} = 1 – {\sin ^2}t = {\cos ^2}t\\
\to \begin{array}{*{20}{l}}
{\smallint \sqrt {1 – {x^2}} {\mkern 1mu} {\rm{d}}x = \smallint \sqrt {{{\cos }^2}t} {\mkern 1mu} \cos t{\mkern 1mu} {\rm{d}}t = \smallint {{\cos }^2}t{\mkern 1mu} {\rm{d}}t = \smallint \frac{{1 + \cos 2t}}{2}{\mkern 1mu} {\rm{d}}t}\\
{ = \smallint \left( {\frac{1}{2} + \frac{1}{2}\cos 2t} \right){\rm{d}}t = \frac{t}{2} + \frac{{\sin 2t}}{4} + C.}
\end{array}
\end{array}\)

(Vì \(
x \in \left[ {0;\frac{\pi }{2}} \right] \Rightarrow \cos x > 0 \Rightarrow \sqrt {{{\cos }^2}x} = \cos x\))

Vậy \(
I = \frac{t}{2} + \frac{{\sin 2t}}{4} + C.\)